If the curves x2-a2-y2-b2-1 and x2-2-y2-2-1

The correct option is A a^2 + b^2 = α^2 + β^2Slope of I = I= b^2 x/a^2 y = m_1 (say) And slope of II = β^x/α^y = m_2 (say) ∴ m_1 m_2 = 1 ∴ b^2 β^2 x^2 = a^2 ...

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Byju's Answer

Standard XII

Mathematics

Angle of Intersection of Two Curves

If the curves...

Question

If the curves $\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1$ and
$\frac{x^{2}}{α^{2}} + \frac{y^{2}}{β^{2}} = 1$

a^{2} + b^{2} = α^{2} + β^{2}

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a^{2} - b^{2} = α^{2} - β^{2}

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a^{2} - b^{2} = α^{2} + β^{2}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

a^{2} + b^{2} = α^{2} - β^{2}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

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Solution

The correct option is A $a^{2} + b^{2} = α^{2} + β^{2}$
Slope of $I = I = \frac{- b^{2} x}{a^{2} y} = m_{1} (s a y)$
And slope of $I I = \frac{β^{x}}{α^{y}} = m_{2} (s a y) ∴ m_{1} m_{2} = - 1 ∴ b^{2} β^{2} x^{2} = a^{2} a^{2} y^{2} . . . (i)$
And $\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} - 1 = 0 \Rightarrow x^{2} (\frac{1}{a^{2}} + \frac{1}{a^{2}}) = - y^{2} (\frac{1}{b^{2}} - \frac{1}{β^{2}}) - - - - (i i)$
From Eqs. (i) and (ii)
$\frac{\frac{1}{a^{2}} - \frac{1}{a^{2}}}{b^{2} β^{2}} = - \frac{(\frac{1}{b^{2}} - \frac{1}{β^{2}})}{a^{2} a^{2}} \Rightarrow α^{2} + β^{2} = a^{2} + b^{2}$

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If the curves x2a2+y2b2=1 and ​x2α2+y2β2=1

The correct option is A a2+b2=α2+β2Slope of I=I=−b2xa2y=m1(say) And slope of II=βxαy=m2(say)∴m1m2=−1∴b2β2x2=a2a2y2...(i) And x2a2+y2b2−1=0⇒x2(1a2+1a2)=−y2(1b2−1β2)−−−−(ii) From Eqs. (i) and (ii) 1a2−1a2b2β2=−(1b2−1β2)a2a2⇒α2+β2=a2+b2

If the curves $\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1$ and
$\frac{x^{2}}{α^{2}} + \frac{y^{2}}{β^{2}} = 1$