The equation of the tangent to the parabola y | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(C) The equation of the parabola is $y^2=16x$,so $4a=16$,which gives $a=4$. The given line is $3x-4y+5=0$,which can be written as $y=\frac{3}{4}x+\fra

Vedclass · Accepted Answer

$4x+3y+9=0$

Vedclass · Answer

(C) The equation of the parabola is $y^2=16x$,so $4a=16$,which gives $a=4$. The given line is $3x-4y+5=0$,which can be written as $y=\frac{3}{4}x+\frac{5}{4}$. The slope of this line is $m_1=\frac{3}{4}$. Since the tangent is perpendicular to this line,its slope $m$ must satisfy $m 	imes m_1 = -1$. Thus,$m = -\frac{4}{3}$. The equation of a tangent to the parabola $y^2=4ax$ with slope $m$ is $y=mx+\frac{a}{m}$. Substituting $a=4$ and $m=-\frac{4}{3}$,we get $y=-\frac{4}{3}x+\frac{4}{-4/3}$. $y=-\frac{4}{3}x-3$. Multiplying by $3$,we get $3y=-4x-9$,which simplifies to $4x+3y+9=0$.

The equation of the tangent to the parabola $y^2=16x$,which is perpendicular to the line $3x-4y+5=0$,is given by

Similar Questions

For the curve $y^2 = 4ax$,the lengths of the tangent,subtangent,normal,and subnormal at the point $(at^2, 2at)$ are respectively:

What is the slope of the normal to the parabola $x^2 + 4y = 0$ at the point $(2, -1)$?

$A$ beam of light travels along the line $y = -4$ from right to left and strikes a parabola at point $P$. If the focus and directrix of the parabola are $(2, 0)$ and $x = -2$ respectively,then find the coordinates of the point where the reflected beam contacts the parabola again.

What is the length of the latus rectum of the parabola $x^2 = -y$?

Vedclass Test Series

Exam Paper Generator

Online Exam Module