Find the equation of the common tangent to th | vedclass

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

Question

(B) The equation of any tangent to the parabola $x^2 = 16y$ is of the form $y = mx - 4m^2$,where $a = 4$ in $x^2 = 4ay$.If this line is also a tangent

Vedclass · Accepted Answer

$x + 2y + 2 = 0$

Vedclass · Answer

(B) The equation of any tangent to the parabola $x^2 = 16y$ is of the form $y = mx - 4m^2$,where $a = 4$ in $x^2 = 4ay$.If this line is also a tangent to the parabola $y^2 = 2x$ (where $a = 1/2$),the condition for tangency $c = a/m$ must be satisfied.Here,$c = -4m^2$ and $a = 1/2$,so $-4m^2 = (1/2)/m$.This simplifies to $m^3 = -1/8$,which gives $m = -1/2$.Substituting $m = -1/2$ into the tangent equation: $y = (-1/2)x - 4(-1/2)^2$.$y = -x/2 - 1$.Multiplying by $2$,we get $2y = -x - 2$,which rearranges to $x + 2y + 2 = 0$.

Find the equation of the common tangent to the parabolas $y^2 = 2x$ and $x^2 = 16y$.

Similar Questions

Let the tangent to the parabola $S: y^{2}=2x$ at the point $P(2,2)$ meet the $x$-axis at $Q$ and the normal at $P$ meet the parabola $S$ at the point $R$. Then the area (in $sq. \ units$) of the triangle $PQR$ is equal to:

$A$ point on the parabola whose focus and vertex are respectively at $\left(\frac{5}{4}, -2\right)$ and $(1, -2)$ is

Let $O$ be the vertex of the parabola $x^{2}=4y$ and $Q$ be any point on it. Let the locus of the point $P$, which divides the line segment $OQ$ internally in the ratio $2:3$, be the conic $C$. Then the equation of the chord of $C$, which is bisected at the point $(1, 2)$, is:

The normals at three points $P, Q, R$ of the parabola $y^2 = 4ax$ meet in $(h, k).$ The centroid of triangle $PQR$ lies on :

The equations of the tangents to the parabola $y^2 = 4ax$ at the ends of its latus rectum are-

Vedclass Test Series

Exam Paper Generator

Online Exam Module