If the curves ax²+by²=1 and cx²+dy²=1 intersect orthogonally, then b-a/d-c=

Correct Answer is : ac⋅bda/c · b/dca⋅db

Correct Answer is : ac⋅bda/c · b/dca⋅db

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TS EAMCET 14-Sep-2020 Shift 1 Solved Paper

Section: Mathematics

Question : 69 of 160

Marks: +1, -0

If the curves

ax^2+by^2=1

cx^2+dy^2=1

\frac{b-a}{d-c}=

Solution:

Given curves

a x^{2}+b y^{2}=1

...(i)

and

\;\; c x^{2}+d y^{2}=1

....(ii)

From Eqs. (i) and (ii),

(a-c) x^{2}+(b-d) y^{2}=0

\;\frac{x^{2}}{y^{2}} = -\left(\frac{b-d}{a-c}\right)

...(iii)

Differentiating Eq. (i) w.r.t.'

x^{'}

, we get

2 a x + 2 b y y^{'} = 0 \Rightarrow y^{'} = -\; \frac{a x}{b y}

Differentiating Eq. (ii) w.r.t'

x^{'}

we get

2 c x + 2 d y y^{'} = 0 \Rightarrow y^{'} = \; \frac{-\alpha x}{d y}

Curves intersect orthogonally

\therefore \;\; \left(\frac{-a x}{b y}\right)\left(\frac{-c x}{d y}\right) = -1 \Rightarrow \;\frac{x^{2}}{y^{2}} = \;\frac{-b d}{a c}

...(iv)

From Eqs. (iii) and (iv), we get

-\left(\frac{b-d}{a-c}\right) = -\;\frac{b d}{a c}

-\left(\frac{b-d}{a-c}\right) = -\;\frac{b d}{a c}

\Rightarrow a b c - a c d = a b d - b c d \Rightarrow b c d - a c d = a b d - a b c

c d (b-a) = a b (d-c) \Rightarrow \;\frac{b-a}{d-c} = \;\frac{a b}{c d}

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