એક ઉપવલય પરનું બિંદુ (4, -1) ને રેખા x + 4y - | vedclass

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Question

$x +4 y -10=0$

$\Rightarrow y =\frac{- x }{4}+\frac{5}{2}$

Let equation of the ellipse be

$\frac{ x ^2}{ a ^2}+\frac{ y ^2}{ b ^2}=1 \quad \l

Vedclass · Accepted Answer

$\frac{{{x^2}}}{{20}}\,\, + \;\,\frac{{{y^2}}}{5}\,\, = \,\,1$

Vedclass · Answer

$x +4 y -10=0$

$\Rightarrow y =\frac{- x }{4}+\frac{5}{2}$

Let equation of the ellipse be

$\frac{ x ^2}{ a ^2}+\frac{ y ^2}{ b ^2}=1 \quad \ldots . . .(2)$

$y = mx + c$ is the tangent to ellipse if $c ^2= a ^2 m ^2+ b ^2$

By $(1)$ we get

$\frac{25}{4}=\frac{a^2}{16}+b^2$

$\Rightarrow 100= a ^2+16 b ^2$

$(4,-1)$ lies on (3)

$\Rightarrow \frac{16}{ a ^2}+\frac{1}{ b ^2}=1$

On solving $(3)$ and $(4)$ we get

$	ext { (3) } \Rightarrow a ^2+16 b ^2=100$

$(4)$ $\Rightarrow a^2+16 b^2=a^2 b^2$

From $(3)$ from $(4)$ we get

$a^2 b^2=100$

$	herefore a ^2=20$ and $b ^2=5$

$\frac{x^2}{20}+\frac{y^2}{5}=1$ is the required equation.

એક ઉપવલય પરનું બિંદુ $(4, -1)$ ને રેખા $x + 4y - 10 = 0$ સ્પર્શેં છે જો તેની અક્ષો યામાક્ષો સાથે સાંપતી હોય, તો તેનું સમીકરણ $(a > b)$

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