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The line $y=x+5$ touches
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Verified Answer
The correct answers are:
the parabola $y^2=20 x$, the ellipse $9 x^2+16 y^2=144$, the hyperbola $\frac{x^2}{29}-\frac{y^2}{4}=1$
(A) $y^2=20 x=4(5) x \quad \therefore$ Tangent $: y=m x+\frac{a}{m}$ for $m=1, a=5$
$\therefore \mathrm{y}=\mathrm{x}+5$ is a tangent to $\mathrm{y}^2=20 \mathrm{x}$
(B) $9 x^2+16 y^2=144$
$c^2=a^2 m^2+b^2 \quad y=x+5 \Rightarrow m=1 ; c=5$
$\Rightarrow \frac{x^2}{16}+\frac{y^2}{9}=1$
$\therefore 5^2=16(1)^2+9 \Rightarrow$ True
$\therefore y=x+5$ is a tangent to $9 x^2+16 y^2=144$
(C) $c^2=a^2 m^2-b^2$
$\frac{x^2}{29}-\frac{y^2}{4}=1 \rightarrow a^2=29 ; b^2=4$
$5^2=29(1)^2-4 \rightarrow$ True
$\therefore \mathrm{y}=\mathrm{x}+5$ is a tangent
(D)
$\therefore \mathrm{y}=\mathrm{x}+5$ is NOT tangent to $\mathrm{x}^2+\mathrm{y}^2=25$
(A), (B), (C) are correct.
$\therefore \mathrm{y}=\mathrm{x}+5$ is a tangent to $\mathrm{y}^2=20 \mathrm{x}$
(B) $9 x^2+16 y^2=144$
$c^2=a^2 m^2+b^2 \quad y=x+5 \Rightarrow m=1 ; c=5$
$\Rightarrow \frac{x^2}{16}+\frac{y^2}{9}=1$
$\therefore 5^2=16(1)^2+9 \Rightarrow$ True
$\therefore y=x+5$ is a tangent to $9 x^2+16 y^2=144$
(C) $c^2=a^2 m^2-b^2$
$\frac{x^2}{29}-\frac{y^2}{4}=1 \rightarrow a^2=29 ; b^2=4$
$5^2=29(1)^2-4 \rightarrow$ True
$\therefore \mathrm{y}=\mathrm{x}+5$ is a tangent
(D)

$\therefore \mathrm{y}=\mathrm{x}+5$ is NOT tangent to $\mathrm{x}^2+\mathrm{y}^2=25$
(A), (B), (C) are correct.
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