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AP and TS EAMCET Engineering Model Paper 18

Question : 26 of 160

Marks: +1, -0

ax+by=1,cx^2+dy^2=1

\frac{a^2}{c}+\frac{b^2}{d}=

Solution:

ax + by =1

\Rightarrow by = 1 - ax

\Rightarrow y = \frac{1- a x}{b}

ax^2+dy^2=1

\Rightarrow cx^2+d(1-ax)^2=1

\Rightarrow b^2cx^2+d(1+a^2x^2-2ax)=b^2

\Rightarrow (b^2c+a^2d)x^2

-2adx+d-b^2=0

both the equations have only one solution ⇒ discriminant = 0

\Rightarrow 4a^2d^2-4(b^2c+a^2d)

(d-b^2)=0

\Rightarrow a^2d^2-(b^2c+a^2d)(d-b^2)=0

\Rightarrow a^2d^2-b^2cd+b^4c

-a^2d^2+a^2b^2d=0

\Rightarrow b^4c+a^2b^2d=b^2cd

\Rightarrow \frac{b^2}{d}+\frac{a^2}{c}=1

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