Show that the curve x^2 a^2+ _1 + y^2 b^2+ _1 =1 and x^2 a^2+ _2 + y^2 b^2+ _2 =1 intersect at right angles.

we have x^2 a^2+ _1 + y^2 b^2+ _1 =1 ...(1) x^2 a^2+ _2 + y^2 b^2+ _2 =1 ...(2) Now we can find the slope of both the curve by differentiating w.r.t.x, 2x a^2+ _1 + 2y dy dx b^2+ _1 =0 and 2x a^2+ _2 + 2y dy dx b^2+ _2 =0 dy dx =- x y b^2+ _1 a^2+ _1 and dy dx =- x y b^2+ _2 a^2+ _2 _1=- x y b^2+ _1 a^2+ _1 and _1=- x y b^2+ _2 a^2+ _2 Substracting (2) from (1), we get, x^2 (1 a^2+ _1 -1 a^2+ _2 )+y^2 (1 b^2+ _1 -1 b^2+ _2 )=0 x^2 y^2 = _2- _1 (b^2+ _1)(b^2+ _2) 1 _1- _2 (e^2+ _1)(e^2+ _2) now, m_1 _2= x^2 y^2 b^2+ _1 a^2+ _1 b^2+ _2 a^2+ _2 = _2- _1 (b^2+ _1)(b^2+ _2) (a^2+ _1)(a^2+ _2) _1- _2 b^2+ _1 a^2+ _1 b^2+ _1 a^2+ _1 =-1 Hence, (1) and (2) cuts orthogonally.

Show that the curve x^2 a^2+ _1 + y^2 b^2+ _1 =1 and x^2 a^2+ _2 …

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Question

Show that the curve $\frac{\text{x}^2}{\text{a}^2+\lambda_1}+\frac{\text{y}^2}{\text{b}^2+\lambda_1}=1$ and $\frac{\text{x}^2}{\text{a}^2+\lambda_2}+\frac{\text{y}^2}{\text{b}^2+\lambda_2}=1$ intersect at right angles.

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Answer

we have $\frac{\text{x}^2}{\text{a}^2+\lambda_1}+\frac{\text{y}^2}{\text{b}^2+\lambda_1}=1\ ...(1)$$\frac{\text{x}^2}{\text{a}^2+\lambda_2}+\frac{\text{y}^2}{\text{b}^2+\lambda_2}=1\ ...(2)$
Now we can find the slope of both the curve by differentiating w.r.t.x,
$\Rightarrow\frac{2\text{x}}{\text{a}^2+\lambda_1}+\frac{2\text{y}\frac{\text{dy}}{\text{dx}}}{\text{b}^2+\lambda_1}=0\text{ and }\frac{2\text{x}}{\text{a}^2+\lambda_2}+\frac{2\text{y}\frac{\text{dy}}{\text{dx}}}{\text{b}^2+\lambda_2}=0$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=-\frac{\text{x}}{\text{y}}\times\frac{\text{b}^2+\lambda_1}{\text{a}^2+\lambda_1}\text{ and }\frac{\text{dy}}{\text{dx}}=-\frac{\text{x}}{\text{y}}\times\frac{\text{b}^2+\lambda_2}{\text{a}^2+\lambda_2}$
$\Rightarrow\text{m}_1=-\frac{\text{x}}{\text{y}}\times\frac{\text{b}^2+\lambda_1}{\text{a}^2+\lambda_1}\text{ and }\Rightarrow\text{m}_1=-\frac{\text{x}}{\text{y}}\times\frac{\text{b}^2+\lambda_2}{\text{a}^2+\lambda_2}$
Substracting (2) from (1), we get,
$\text{x}^2\Big(\frac{1}{\text{a}^2+\lambda_1}-\frac{1}{\text{a}^2+\lambda_2}\Big)+\text{y}^2\Big(\frac{1}{\text{b}^2+\lambda_1}-\frac{1}{\text{b}^2+\lambda_2}\Big)=0$
$\Rightarrow\frac{\text{x}^2}{\text{y}^2}=\frac{\lambda_2-\lambda_1}{(\text{b}^2+\lambda_1)(\text{b}^2+\lambda_2)}\times\frac{1}{\frac{\lambda_1-\lambda_2}{(\text{e}^2+\lambda_1)(\text{e}^2+\lambda_2)}}$
now,
$\text{m}_1\times\text{m}_2=\frac{\text{x}^2}{\text{y}^2}\times\frac{\text{b}^2+\lambda_1}{\text{a}^2+\lambda_1}\times\frac{\text{b}^2+\lambda_2}{\text{a}^2+\lambda_2}$
$=\frac{\lambda_2-\lambda_1}{(\text{b}^2+\lambda_1)(\text{b}^2+\lambda_2)}\times\frac{(\text{a}^2+\lambda_1)(\text{a}^2+\lambda_2)}{\lambda_1-\lambda_2}\times\frac{\text{b}^2+\lambda_1}{\text{a}^2+\lambda_1}\times\frac{\text{b}^2+\lambda_1}{\text{a}^2+\lambda_1}$
$=-1$
Hence, (1) and (2) cuts orthogonally.