Tangent to Parabola \(y^2=4ax\) at \( (x_1, y_1) \)
Let the point of tangency be \( (x_1, y_1) \) on the parabola \(y^2=4ax\), we can write that \( y_1^2=4ax_1 \) (i)
Now the equation of tangent to the parabola at \( (x_1, y_1) \) is \( y-y_1 = m(x-x_1) \) (A)
The slope of the tangent to the parabola is \( y^2 = 4ax \)
Differentiate both sides with respect to \( x \), we get \( 2y \frac{dy}{dx} = 4a \)
or \( \frac{dy}{dx} = \frac{4a}{2y} = \frac{2a}{y} \)
Now, the equation of the tangent line at a point \( (x_1, y_1) \) with slope \( m \) is \( y - y_1 = m(x - x_1) \)
or \( y - y_1 = \frac{2a}{y_1}(x - x_1) \) Answer
or \( y y_1 - y_1^2 = 2a(x - x_1) \)
or \( y y_1 - 4ax_1 = 2ax - 2ax_1 \) From (i): \( y_1^2=4ax_1 \)
or \( y y_1 = 2ax + 2ax_1 \)
or \( y y_1 = 2a(x + x_1) \)
NOTE
Now, the equation of normal is \( y - y_1 = -\frac{y_1}{2a}(x - x_1) \)
Tangent to Parabola \(x^2=4ay\) at \( (x_1, y_1) \)
Let the point of tangency be \( (x_1, y_1) \) on the parabola \( x^2 = 4ay \), we can write that \( x_1^2 = 4ay_1 \) (i)
Now the equation of tangent to the parabola at \( (x_1, y_1) \) is \( y - y_1 = m(x - x_1) \) (A)
The slope of the tangent to the parabola is \( x^2 = 4ay \)
Differentiate both sides with respect to \( x \), we get \( 4a \frac{dy}{dx} =2x \)
or \( \frac{dx}{dy} = \frac{2x}{4a} = \frac{x}{2a} \)
Now, the equation of the tangent line at a point \( (x_1, y_1) \) with slope \( m \) is \( y - y_1 = m(x - x_1) \)
or \( y - y_1 = \frac{x_1}{2a}(x - x_1) \) Answer
or \( 2a(y - y_1) = xx_1 - x_1^2) \)
or \( 2ay - 2ay_1 = xx_1 - 4ay_1) \) From (i): \( x_1^2 = 4ay_1 \)
or \( 2ay - 2ay_1 = xx_1 - 4ay_1) \)
or \( x x_1 = 2a(y + y_1) \)
NOTE
Now, the equation of normal is \( y - y_1 = -\frac{2a}{x_1}(x - x_1) \)
Condition of Tangentcy: line \( y = mx + c \) to the parabola \( y^2 = 4ax \)
Let \( y = mx + c \) be a tangent to the parabola \( y^2 = 4ax \), then they must intersect at exactly one point.
Substituting \( y = mx + c \) on \( y^2 = 4ax \), we get \( (mx + c)^2 = 4ax \) \( m^2x^2 + 2mcx + c^2 = 4ax \) \( m^2x^2 + (2mc - 4a)x + c^2 = 0 \)
This is a quadratic equation in \( x \), this quadratic equation must have exactly one solution, so the discriminant must be zero, thus we have \( b^2 =4ac\)
or\( (2mc - 4a)^2 = 4m^2c^2 \)
or\( 2mc - 4a = \pm 2mc \) (A)
Taking positive sign from (A), we get \( 2mc - 4a = 2mc \)
or \( -4a = 0 \)
or \( a = 0 \) (not possible as \( a \neq 0 \) for a parabola)
Taking negative sign from (A), we get \( 2mc - 4a = -2mc \)
or\( 4mc = 4a \)
or \( mc = a \)
Therefore, the condition for the line \( y = mx + c \) to be tangent to the parabola \( y^2 = 4ax \) is \( mc = a \) or \(c=\frac{a}{m} \)
NOTE
Equation of tangent:
The equation of tangent line to the parabola \( y^2 = 4ax \) is \( y = mx + c \)
or\( y = mx + \frac{a}{m} \) Answer
The point of contact:
The point of contact of tangent \( y = mx + \frac{a}{m} \) to the parabola \( y^2 = 4ax \) is \( (x_1,y_1) \)
Then \( y = mx + \frac{a}{m} \) (A) \( yy_1 = 2a(x + x_1) \) (B)
must be identical
So, we can write that \( \frac{1}{y_1}= \frac{m}{2a}=\frac{a/m}{2ax_1} \)
Equationg first and second part gives \( \frac{1}{y_1}= \frac{m}{2a} \) or \(y_1=\frac{2a}{m}\)
Equationg second and third part gives \( \frac{m}{2a}=\frac{a/m}{2ax_1} \) or \(x_1=\frac{a}{m^2}\)
Hence, the point of contact is \( (x_1,y_1)=(\frac{a}{m^2},\frac{2a}{m}) \) Answer
Condition of Tangentcy: line \( y = mx + c \) to the parabola \( x^2 = 4ay \)
Let \( y = mx + c \) be a tangent to the parabola \( x^2 = 4ay \), then they must intersect at exactly one point.
Substituting \( y = mx + c \) into \( x^2 = 4ay \), we get \( x^2 = 4a(mx + c) \) \( x^2 = 4amx + 4ac \) \( x^2 - 4amx - 4ac = 0 \)
This is a quadratic equation in \( x \), which must have exactly one solution, so the discriminant must be zero, thus we have \( b^2 = 4ac \)
or\( (4am)^2 - 4(1)(-4ac) = 0 \)
or\( 16a^2m^2 + 16ac = 0 \)
or\( 16a(am^2 + c) = 0 \)
Since \( a \neq 0 \) for a parabola, we have: \( am^2 + c = 0 \)
or\( c = -am^2 \)
Therefore, the condition for the line \( y = mx + c \) to be tangent to the parabola \( x^2 = 4ay \) is \( c = -am^2 \)
NOTE
Equation of tangent:
The equation of tangent line to the parabola \( x^2 = 4ay \) is \( y = mx + c \)
or\( y = mx -am^2 \)
The point of contact:
The point of contact of tangent \( y = mx -am^2 \) to the parabola \( x^2 = 4ay \) is \( (x_1,y_1) \)
Then \( y = mx -am^2 \) (A) \( x x_1 = 2a(y + y_1) \) (B)
must be identical
So, we can write that \( \frac{1}{2a}= \frac{m}{x_1}=\frac{am^2}{2ay_1} \)
Equationg first and second part gives \( \frac{1}{2a}= \frac{m}{x_1} \) or \(x_1=\frac{m}{2a}\)
Equationg first and third part gives \(\frac{1}{2a}=\frac{am^2}{2ay_1} \) or \(y_1=am^2\)
Hence, the point of contact is \( (x_1,y_1)=(\frac{m}{2a},am^2) \)
Normal to Parabola \(y^2=4ax\) at \( (x_1, y_1) \)
At \( (x_1, y_1) \), the equation of normal is \( y-y_1 = m(x-x_1) \) (A)
The slope of the tangent to the parabola is \( y^2 = 4ax \)
or \( 2y \frac{dy}{dx} = 4a \)
or \( \frac{dy}{dx} = \frac{4a}{2y} = \frac{2a}{y} \)
The slope of the normal to the parabola \( y^2 = 4ax \) at \( (x_1, y_1) \) is \( -\frac{y_1}{2a} \)
Now, the equation of the normal line at a point \( (x_1, y_1) \) to the parabola \( y^2 = 4ax \) is \( y - y_1 = m(x - x_1) \)
or \( y - y_1 = -\frac{y_1}{2a}(x - x_1) \)
Normal to Parabola \(y^2=4ax\) in Slope form
The equation of the normal to the parabola \( y^2 = 4ax \) is \( y - y_1 = m(x - x_1) \)
or \( y - y_1 = -\frac{y_1}{2a}(x - x_1) \)
where \(m = -\frac{y_1}{2a} \)
So, we have \(y_1=-2am \) and \(x_1=\frac{y_1^2}{4a}=\frac{(-2am)^2}{4a}= am^2 \)
Hence the equation of normal is \( y - y_1 = -\frac{y_1}{2a}(x - x_1) \)
or \( y - (-2am ) = m[x - (am^2)] \)
or \( y =mx-2am -am^3 \)
Normal to Parabola \(x^2=4ay\) at \( (x_1, y_1) \)
At \( (x_1, y_1) \), the equation of normal is \( y-y_1 = m(x-x_1) \) (A)
The slope of the tangent to the parabola is \( x^2 = 4ay \)
Differentiate both sides with respect to \( x \), we get \( 4a \frac{dy}{dx} =2x \)
or \( \frac{dx}{dy} = \frac{2x}{4a} = \frac{x}{2a} \)
The slope of the normal to the parabola \( x^2 = 4ay \) at \( (x_1, y_1) \) is \( -\frac{2a}{x_1} \)
Now, the equation of the normal line at a point \( (x_1, y_1) \) to the parabola \( x^2 = 4ay \) is \( y - y_1 = m(x - x_1) \)
or \( y - y_1 = -\frac{2a}{x_1} (x - x_1) \)
Normal to Parabola \(x^2=4ay\) in Slope form
The equation of the normal to the parabola \( x^2 = 4ay \) is \( y - y_1 = m(x - x_1) \)
or \( y - y_1 = -\frac{2a}{x_1}(x - x_1) \)
where \(m =-\frac{2a}{x_1} \)
So, we have \(x_1=-\frac{2a}{m} \) and \(y_1=\frac{x_1^2}{4a}=\frac{4a^2}{m^24a}= \frac{a}{m} \)
Hence the equation of normal is \( y - y_1 = -\frac{y_1}{2a}(x - x_1) \)
or \( y - (-\frac{2a}{m} ) = m[x - (\frac{a}{m} )] \)
or \( y =mx-\frac{2a}{m}-a \)
Tangent and Normal to Parabola \(y^2=4ax\) in parametric form
Let \(y^2=4ax\) be a parabola in which parametric coordinate is \((at^2,2at)\) \( y_1^2=4ax_1 \) (i)
Now the equation of tangent to the parabola is \( y-y_1 = m(x-x_1) \) (A)
The slope of the tangent to the parabola is \( m= \frac{2a}{y} =\frac{2a}{2at}=\frac{1}{t}\)
Now, the equation of the tangent is \( y - y_1 = m(x - x_1) \)
or \( y - y_1 = \frac{1}{t}(x - x_1) \)
Now, the equation of normal is \( y - y_1 = m(x - x_1) \)
or \( y - y_1 = -t(x - x_1) \)
Exercise
Find the equation of tangent and normal to the parabola
Equation of the Tangent
The parabola is \( y^2 = 4ax \), so we have \( 4a = 8 \), so \( a = 2 \)
Also, the slope of tangent to the parabola at \( (x_1, y_1)=(2,-4) \) is \( m_1 = \frac{2a}{y_1} =\frac{2.2}{-4}=-1\)
Now, equation of tangent to the parabola is \( y-y_1=m_1(x-x_1)\) \( y+4 = (-1) (x - 2) \)
or\( y+4=-x+2 \)
or\( x + y + 2 = 0 \) Answer: tangent line
Equation of the Normal
Now, slope of normal to the parabola at \( (x_1, y_1)=(2,-4) \) is \( m_2 = \frac{y_1}{2a} =1\)
Now, equation of normal to the parabola is \( y-y_1=m_2(x-x_1)\) \( y+4 = (1) (x - 2) \)
or\( y+4=x-2 \)
or\( x - y -6= 0 \) Answer: normal line
Equation of the Tangent
The parabola is \( x^2 = 4ay \), so we have \( 4a = 12 \), so \( a = 3 \)
Also, the slope of tangent to the parabola at \( (x_1, y_1) = (-6, 3) \) is \( m_1 = \frac{x_1}{2a} =\frac{-6}{2 \cdot 3} = -1 \)
Now, the equation of the tangent to the parabola is \( y - y_1 = m_1(x - x_1) \) \( y - 3 = (-1) (x + 6) \)
or\( y - 3 = -x - 6 \)
or\( x + y + 3 = 0 \) Answer: tangent line
Equation of the Normal
Now, the slope of the normal to the parabola at \( (x_1, y_1) = (-6, 3) \) is \( m_2 = -\frac{1}{m_1} = 1 \)
Now, the equation of the normal to the parabola is \( y - y_1 = m_2(x - x_1) \) \( y - 3 = (1) (x + 6) \)
or\( y - 3 = x + 6 \)
or\( x - y + 9 = 0 \) Answer: normal line
Equation of the Tangent
The parabola is \( x^2 = 4y \), so we have \( 4a = 4 \), so \( a = 1 \)
Given \( x_1 = 6 \), we find the corresponding \( y_1 \) by substituting into the equation: \( y_1 = \frac{x_1^2}{4} = \frac{6^2}{4} = 9 \)
So, the point of tangency is \( (6, 9) \).
Now, the slope of the tangent to the parabola at \( (x_1, y_1) = (6, 9) \) is \( m_1 = \frac{x_1}{2a} = \frac{6}{2 \cdot 1} = \frac{6}{2} =3 \)
Now, the equation of the tangent to the parabola is \( y - y_1 = m_1(x - x_1) \) \( y - 9 = 3(x - 6) \)
or\( y - 9 = 3x - 18 \)
or\( 3x - y -9 = 0 \) Answer: tangent line
Equation of the Normal
Now, the slope of the normal to the parabola at \( (x_1, y_1) = (6, 9) \) is \( m_2 = -\frac{1}{m_1} = -\frac{1}{3} \)
Now, the equation of the normal to the parabola is \( y - y_1 = m_2(x - x_1) \) \( y - 9 = -\frac{1}{3}(x - 6) \)
or\( 3y - 27 = -x + 6 \)
or\( x +3 y - 33 = 0 \) Answer: normal line
Equation of the Tangent
The parabola is \( y^2 = 16ax \).
Given \( y_1 = -4a \), we find the corresponding \( x_1 \) by substituting into the equation: \( (-4a)^2 = 16a \cdot x_1 \implies 16a^2 = 16a \cdot x_1 \implies x_1 = a \)
So, the point of tangency is \( (a, -4a) \).
Now, the slope of the tangent to the parabola at \( (x_1, y_1) = (a, -4a) \) is \( m_1 = \frac{8a}{y_1} = \frac{8a}{-4a} = -2 \)
Now, the equation of the tangent to the parabola is \( y - y_1 = m_1(x - x_1) \) \( y + 4a = -2(x - a) \)
or\( y + 4a = -2x + 2a \)
or\( 2x + y + 2a = 0 \) Answer: tangent line
Equation of the Normal
The slope of the normal to the parabola at \( (x_1, y_1) = (a, -4a) \) is \( m_2 = \frac{-1}{m_1} = \frac{1}{2} \)
Now, the equation of the normal to the parabola is \( y - y_1 = m_2(x - x_1) \) \( y + 4a = \frac{1}{2}(x - a) \)
or\( 2(y + 4a) = x - a \)
or\( x - 2y - 9a = 0 \) Answer: normal line
We know that, end of the latus rectum are \(a,\pm 2a)\).
For the parabola \( y^2 = 12x \), we have \( 4a = 12 \), so \( a = 3 \). The end of latus rectum are \( (3, 6) \) and \( (3, -6) \).
At the End \( (3, 6) \) Equation of the Tangent
The slope of the tangent to the parabola \( y^2 = 12x \) at \( (x_1, y_1) = (3, 6) \) is \( m_1 = \frac{y_1}{2a} = \frac{6}{6} = 1 \)
Now, the equation of the tangent is \( y - y_1 = m_1(x - x_1) \) \( y - 6 = 1(x - 3) \)
or\( y - 6 = x-3 \)
or\( x-y+3=0 \) Answer: tangent line Equation of the Normal
The slope of the normal is the negative reciprocal of the tangent’s slope: \( m_2 = -\frac{1}{m_1} = -1 \)
Now, the equation of the normal is \( y - y_1 = m_2(x - x_1) \) \( y - 6 = -1(x - 3) \)
or\( y - 6 = -x + 3 \)
or\( x + y - 9 = 0 \) Answer: normal line
At the End \( (3, -6) \)
The slope of the tangent to the parabola \( y^2 = 12x \) at \( (x_1, y_1) = (3, -6) \) is \( m_1 = \frac{y_1}{2a} = \frac{6}{-6} = -1 \)
Now, the equation of the tangent is \( y - y_1 = m_1(x - x_1) \) \( y + 6 = -1(x - 3) \)
or\( y + 6 = -x+3 \)
or\( x+y+3=0 \) Answer: tangent line Equation of the Normal
The slope of the normal is the negative reciprocal of the tangent’s slope: \( m_2 = -\frac{1}{m_1} = 1 \)
Now, the equation of the normal is \( y - y_1 = m_2(x - x_1) \) \( y + 6 = 1(x - 3) \)
or\( y + 6 = x - 3 \)
or\( x-y -9 = 0 \) Answer: normal line
Solve the followings
Prove that the line \(3x+4y+6=0\) is tangent to the parabola \(2y^2=9x\) and find its point of contact
The given parabola is \(2y^2 = 9x\), which is \( y^2 = \frac{9}{2} x \) (i)
Thus, \( a = \frac{9}{8} \)
The given line is \(3x + 4y + 6 = 0\)
or \( y = -\frac{3}{4}x - \frac{3}{2} \) (ii)
The condition for line (ii) to be tangent to the parabola (i) is \(c=\frac{a}{m}\) \(- \frac{3}{2} =\frac{\frac{9}{8}}{ -\frac{3}{4}}\) TRUE
So, the line (ii) is tangent to the parabola (i)
Next, to find the point of contact, we substitute \(x = -\frac{4y + 6}{3}\) into the parabola , so that \( 2y^2 = 9 \left(-\frac{4y + 6}{3}\right) \) \( 2y^2 = -12y - 18 \) \( 2y^2 + 12y + 18 = 0 \) \( y = -3 \)
Substitute \( y = -3 \) into the line equation \(3x + 4y + 6 = 0\) , we get \( 3x + 4(-3) + 6 = 0 \) \( x=2 \)
So, the point of contact is \( (2, -3) \)
Prove that the line \(lx+my+n=0\) touches the parabola \(y^2=4ax\) if \(ln =am^2\)
The given parabola is \(y^2=4ax\) (i)
The given line is \(lx+my+n=0\)
or \( y = -\frac{l}{m}x - \frac{n}{m} \) (ii)
The condition for line (ii) to be tangent to the parabola (i) is \(c=\frac{a}{m}\) \(- \frac{n}{m} =\frac{a}{ -\frac{l}{m}}\) TRUE \(ln =am^2\)
For what value of a, will the straight line \(y=2x+3\) touch the parabola \(y^2=4ax\)
The given parabola is \(y^2=4ax\) (i)
The given line is \(y=2x+3\) (ii)
The condition for line (ii) to be tangent to the parabola (i) is \(c=\frac{a}{m}\) \(3 =\frac{a}{2}\) TRUE \(a=6\)
If the line \(2x+4y=3\) touches the parabola \(y^2=4ax\) find the length of the latus rectum
The given parabola is \(y^2=4ax\) (i)
The given line is \(2x+4y=3\) \(y=-\frac{1}{2}x+\frac{3}{4}\) (ii)
The condition for line (ii) to be tangent to the parabola (i) is \(c=\frac{a}{m}\) \(\frac{3}{4} =\frac{a}{-\frac{1}{2}}\) TRUE \(a=-\frac{3}{8}\)
Now, the length of rectum is \(4a=4 \times \frac{3}{8}=\frac{3}{2} \)
Solve the followings
Find the equation of tangent to the parabola \(y^2=4x\)
(i) parallel to the line \(x-2y+6=0\).
(ii) perpendicular to the line \(2x-y=4\)
Also find the point of contact.
(i) parallel to the line \(x-2y+6=0\). Also find the point of contact.
The given parabola is \(y^2=4x\) (i)
Thus, \( 4a = 4\) gives \(a=1\)
The equation of tangent to parabola, which is parallel to \( x - 2y + 6 = 0 \) is \( x - 2y + k = 0 \) (A) \( 2y = x + k \) \( y = \frac{1}{2}x +\frac{k}{2} \) (ii)
The condition for the line (ii) to be tangent to the parabola (i) is \(c=\frac{a}{m}\) \(\frac{k}{2} =\frac{1}{\frac{1}{2}}\) \(k=4\)
Hence, the required equation of tangent line is, from (A) \( x - 2y + 4 = 0 \)
Next, to find the point of contact, we substitute \(x = 2y - 4\) into the parabola , so that \( y^2 = 4 \left(2y - 4 \right) \) \( y^2 = 8y - 16 \) \( y^2 - 8y +16 = 0 \) \( y = 4 \)
Substitute \( y = 4 \) into the line equation \(x - 2y + 4 = 0\) , we get \( x - 2(4) + 4 = 0\) \( x=4 \)
So, the point of contact is \( (4,4) \)
(ii) perpendicular to the line \(2x-y=4\) Also find the point of contact.
The given parabola is \(y^2=4x\) (i)
Thus, \( 4a = 4\) gives \(a=1\)
The equation of tangent to parabola, which is perpendicular to \(2x-y=4\) is \( x + 2y + k = 0 \) (A) \( 2y = -x - k \) \( y = -\frac{1}{2}x -\frac{k}{2} \) (ii)
The condition for the line (ii) to be tangent to the parabola (i) is \(c=\frac{a}{m}\) \(-\frac{k}{2} =\frac{1}{-\frac{1}{2}}\) \(k=4\)
Hence, the required equation of tangent line is, from (A) \( x + 2y + 4 = 0 \)
Next, to find the point of contact, we substitute \(x = - 2y - 4\) into the parabola , so that \( y^2 = 4 \left(-2y - 4 \right) \) \( y^2 = -8y - 16 \) \( y^2 + 8y +16 = 0 \) \( y =- 4 \)
Substitute \( y = -4 \) into the line equation \(x + 2y + 4 = 0\) , we get \( x + 2(-4) + 4 = 0\) \( x=4 \)
So, the point of contact is \( (4,-4) \)
Find the rquation of normal line to the parabola \(y^2=3x\)
(i) parallel to the line \(y=2x+1\)
(ii) perpendicular to the line \(3x+2y=8\)
The given parabola is \(y^2=3x\) (i)
Thus, \( 4a = 3\) gives \(a=\frac{3}{4}\)
The equation of normal to parabola, which is parallel to \(y=2x+1\) is \( y=2x+k \) (A)
The condition for the line (A) to be normal to the parabola (i) is \(c=-2am-am^3\) \(k=-2\frac{3}{4}. 2-\frac{3}{4} (8)\) \(k=-9\)
Hence, the required equation of tangent line is, from (A) \( y=2x-9 \)
(ii) perpendicular to the line \(3x+2y=8\)
The given parabola is \(y^2=3x\) (i)
Thus, \( 4a = 3\) gives \(a=\frac{3}{4}\)
The equation of normal to parabola, which is perpendicular to \(3x+2y=8\) is \( 2x-3y+k=0 \) (A)\( 3y=2x+k \) \( y=\frac{2}{3}x+\frac{k}{3} \) (ii)
The condition for the line (ii) to be normal to the parabola (i) is \(c=-2am-am^3\) \(\frac{k}{3} =-2\frac{3}{4}. \frac{2}{3} -\frac{3}{4} (\frac{8}{27})\) \(k=-\frac{11}{3}\)
Hence, the required equation of tangent line is, from (A) \( 3y=2x-\frac{11}{3} \) \( 9y=6x-11\)
Solve the followings
Show that tangent to the parabola \(y^2=4x\) and \(x^2=4y\) at (1,2) and (-2,1) respectively are at right angles.
Slope of the tangent to the Parabola \( y^2 = 4x \) at \( (1, 2) \)is \( m_1 = \frac{y_1}{2a} \) \( m_1 = \frac{2}{2} = 1 \)
Next, slope of the tangent to the Parabola \( x^2 = 4y \) at \( (-2, 1) \)is \( m_2 = \frac{x_1}{2a} \) \( m_2 = \frac{-2}{2} = -1 \)
The tangents are perpendicular if the product of their slopes is \(-1\). \( m_1 \times m_2 = 1 \times -1 = -1 \)
Since the product of the slopes is \(-1\), the tangents are at right angles
Show that normal to the parabola \(y^2=8x\) at (2,4) meets the parabola again in (18,-12)
The equation of the parabola is \( y^2 = 8x \), so the slope of normal to the parabola is \( m= -\frac{2a}{y_1} \) \( m= - \frac{2.2}{4} = -1 \)
Now, equation of the Normal at \( (2, 4) \) is \( y - y_1 = m(x - x_1) \) \( y - 4 = -1(x - 2) \) \( y = -x + 6 \) \( x + y - 6 = 0 \)
To find where the normal meets the parabola again, substitute the equation of the normal \( x + y - 6 = 0 \) into the equation of the parabola \( y^2 = 8x \), we get \( y^2 = 8x \) \( (6 - x)^2 = 8x \) \( 36 - 12x + x^2 = 8x \) \( x^2 - 20x + 36 = 0 \) \( x = 2 \) or \( x = 18 \)
When \( x = 2 y = 4 \) is the original point, and when \( x = 18 \), we get \( y = 6 - 18 = -12 \).
Therefore, the normal meets the parabola again at \( (18, -12) \)
Solve the followings
Find the equation of tangent to the parabola \(y^2=6x\) making angle 45 degree with the x-axis. Also find the point of contact.
Given that, parabola is \(y^2=6x\), thus, \( 4a = 6\) gives \(a=\frac{3}{2}\)
Also, given that tangent makes an angle of 45 degrees with the x-axis. Therefore, the slope \(m\) of the tangent is \( m = \tan(45^\circ) = 1 \)
Now, the equation of the tangent to the parabola \( y^2 = 4ax \) is \( y=mx+\frac{a}{m}\) \( y = 1.x + \frac{3}{2} \) \( y = x + \frac{3}{2} \)
To find the point of contact, we substitute the tangent equation \( y = x +\frac{3}{2}\) into the parabola equation \( y^2 = 6x \), so we get \( y^2 = 6x \) \( (x + \frac{3}{2})^2 = 6x \) \( (2x + 3)^2 = 24x \) \( (2x - 3)^2 = 0 \) \( x= \frac{3}{2}\) gives \( y = \frac{3}{2} + \frac{3}{2} =3\)
Hence, the point of contact is \( (\frac{3}{2},3)\).
A tangent to the parabola \(y^2=12x\) makes an angle 45 degree with the straight line \(2y=x+3\). Find its equation and point of contact.
Find the equation of tangents to the parabola \(y^2=12x\) drawn through the points (-1,2). Also find the point of contact and the angle between two tangents.
Find the equation of common tangents to the parabolas \( y^2=4x\) and \( x^2=4y\)
Solve the followings
Show that the pair of tangents drawn from the point (-2,3) to the parabola \(y^2=8x\) are at right angle
Prove that the tangents at the extrimities of the latus rectum of a parabola \(y^2=16x\) are at right angles
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