Let the tangent to the parabola y^2=12x at the point (3, ) be perpendicular to the line 2x…

Question

Let the tangent to the parabola ${y}^{2}=12x$ at the point $(3,\alpha )$ be perpendicular to the line $2x+2y=3$. Then the square of distance of the point $(6,-4)$ from the normal to the hyperbola ${\alpha }^{2}{x}^{2}-9{y}^{2}=9{\alpha }^{2}$ at its point $(\alpha -1,\alpha +2)$ is equal to \_\_\_\_\_\_\_\_\_.

Answer: 116

Accepted Answer

$\because P(3,\alpha )$ lies on ${y}^{2}=12x$

$\Rightarrow \alpha =\pm 6$

${\frac{dy}{dx}|}_{(3,\alpha )}=\frac{6}{\alpha }=1\Rightarrow \alpha =6(\alpha =-6)$

Now, hyperbola $\frac{{x}^{2}}{9}-\frac{{y}^{2}}{36}=1$, normal at

$\begin{gathered} Q(\alpha -1,\alpha +2)\frac{9x}{5}+\frac{36y}{8}=45 \ \Rightarrow 2x+5y-50=0 \end{gathered}$

Now, distance of $(6,-4)$ from $2x+5y-50=0$ is equal to

$\begin{gathered} |\frac{2(6)-5(4)-50}{\sqrt{{2}^{2}+{5}^{2}}}|=\frac{58}{\sqrt{29}} \ \Rightarrow =116 \end{gathered}$

Solutions · 1