If two normals to the parabola y^2 = 4x passi | vedclass

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

Question

(A) The equation of a normal to the parabola $y^2 = 4ax$ with slope $m$ is $y = mx - 2am - am^3$. Here $a = 1$,so the equation is $y = mx - 2m - m^3$.

Vedclass · Accepted Answer

$y = x - 3$

Vedclass · Answer

(A) The equation of a normal to the parabola $y^2 = 4ax$ with slope $m$ is $y = mx - 2am - am^3$. Here $a = 1$,so the equation is $y = mx - 2m - m^3$.Since the normal passes through $(15, 12)$,we have $12 = 15m - 2m - m^3$,which simplifies to $m^3 - 13m + 12 = 0$.Let the slopes of the three normals be $m_1, m_2, m_3$. From the cubic equation,the sum of the roots is $m_1 + m_2 + m_3 = 0$.Given the two normals $4x + y = 72$ (slope $m_1 = -4$) and $3x - y = 33$ (slope $m_2 = 3$),we have $-4 + 3 + m_3 = 0$,which gives $m_3 = 1$.The equation of the third normal with slope $m = 1$ passing through $(15, 12)$ is $y - 12 = 1(x - 15)$,which simplifies to $y = x - 3$.

If two normals to the parabola $y^2 = 4x$ passing through the point $(15, 12)$ are $4x + y = 72$ and $3x - y = 33$,find the third normal.

Similar Questions

At which point does the line $x = my + \frac{a}{m}$ touch the parabola $x^2 = 4ay$?

Two parabolas have the same focus $(4, 3)$ and their directrices are the $x$-axis and the $y$-axis,respectively. If these parabolas intersect at the points $A$ and $B$,then $(AB)^2$ is equal to

Let $x+y=k$ be a normal to the parabola $y^2=12x$. If $p$ is the length of the perpendicular from the focus of the parabola onto this normal,then $4k-2p^2$ is equal to

Two perpendicular tangents to the parabola $y^2 = 4ax$ always intersect on the line:

On the parabola $y = x^2$,the point at the least distance from the straight line $y = 2x - 4$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module