"Let the tangent drawn to the parabola $y^{2}=24 x$ at the point $(\alpha, \beta)$ is perpendicular to the line $2 x+2 y=5$. Then the normal to the hyperbola $\frac{x^{2}}{\alpha^{2}}-\frac{y^{2}}{\beta^{2}}=1$ at the point $(\alpha+4, \beta+4)$ does NOT pass through the point :"

Question

Accepted Answer

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Any tangent to ${y^2} = 24x$ at ($\alpha$, $\beta$)

$\beta y = 12(x + \alpha )$

Slope $= {{12} \over \beta }$ and perpendicular to $2x + 2y = 5$

$\Rightarrow {{12} \over \beta } = 1 \Rightarrow \beta = 12,\,\alpha = 6$

Hence hyperbola is ${{{x^2}} \over {36}} - {{{y^2}} \over {144}} = 1$ and normal is drawn at (10, 16)

Equation of normal ${{36\,.\,x} \over {10}} + {{144\,.\,y} \over {16}} = 36 + 144$

$\Rightarrow {x \over {50}} + {y \over {20}} = 1$

This does not pass though (15, 13) out of given option.

"

Let the tangent drawn to the parabola $y^{2}=24 x$ at the point $(\alpha, \beta)$ is perpendicular to the line $2 x+2 y=5$. Then the normal to the hyperbola $\frac{x^{2}}{\alpha^{2}}-\frac{y^{2}}{\beta^{2}}=1$ at the point $(\alpha+4, \beta+4)$ does NOT pass through the point :

Solution

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Let the tangent drawn to the parabola $y^{2}=24 x$ at the point $(\alpha, \beta)$ is perpendicular to the line $2 x+2 y=5$. Then the normal to the hyperbola $\frac{x^{2}}{\alpha^{2}}-\frac{y^{2}}{\beta^{2}}=1$ at the point $(\alpha+4, \beta+4)$ does NOT pass through the point :

Solution

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