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  • If the curves  <img alt="\mathrm {\frac{x^2}{\alpha}+\frac{y^2}{4}=1}" src="https://entrancecorner.oncodecogs.com/gif.latex?%5Cmathrm%20%7B%5Cfrac%7Bx%5E2%7D%7B%5Calpha%7D&plus;%5Cfrac%7By

If the curves  \mathrm {\frac{x^2}{\alpha}+\frac{y^2}{4}=1}  and   \mathrm {y^2=16 x}  intersect at right angles, then the value of \mathrm { \alpha} is

Option: 1

1


Option: 2

2


Option: 3

5


Option: 4

4


Answers (1)

best_answer

Let  \mathrm {P\left(x_1, y_1\right)} be a common point
Tangents to the curves at  \mathrm {P } are
\mathrm { y y_1=8\left(x+x_1\right) \text { and } \frac{x x_1}{\alpha}+\frac{y y_1}{4}=1 } 
Product of the slopes \mathrm { =\frac{8}{y_1} \times\left(\frac{-4 x_1}{\alpha y_1}\right)=-1 } 
\mathrm { \Rightarrow y_1^2=\frac{32}{\alpha} x_1 }......\mathrm{(i)} 
As  \mathrm { P } lies on the parabola \mathrm { y^2=16 x } 
\mathrm { \Rightarrow y_1^2=16 x_1 }......\mathrm{(ii)}
From (i) and (ii) we get \mathrm { \alpha=2. }

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rishi.raj

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