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  • The angle between a pair of tangents drawn from a point to the parabola <img alt="\mathrm{y^2=4 a x\:

The angle between a pair of tangents drawn from a point \mathrm{P} to the parabola \mathrm{y^2=4 a x\: is\: 45^{\circ}}. Then the locus of the point \mathrm{P} is
 

Option: 1

\mathrm{(x+2 a)^2-y^2=8 a^2}
 


Option: 2

\mathrm{2 x^2-y^2=8 a^2}
 


Option: 3

\mathrm{(x+4 a)^2-y^2=8 a^2}
 


Option: 4

\mathrm{(x+3 a)^2-y^2=8 a^2}


Answers (1)

best_answer

Let \mathrm{P(\alpha, \beta)} be any point on the locus. Equation of the pair of tangents from \mathrm{P(\alpha, \beta)} to the parabola \mathrm{y^2=4 a x} is

\mathrm{ [\beta y-2 a(x+\alpha)]^2=\left(\beta^2-4 a \alpha\right)\left(y^2-4 a x\right)\left(\because T^2=S S_1\right) }.....(1)

\mathrm{ Now, a= coefficient \: of \: x^2=4 a^2 }

\mathrm{2 h= coefficient \: of \: x y=-4 a \beta}

\mathrm{and \: b= coefficient \: of\: y^2=\beta^2-\left(\beta^2-4 a \alpha\right)=4 a \alpha}

\mathrm{\text{Since the angle between the two lines of equation }(1) \: is \: \: 45^{\circ},}

\mathrm{ \text { we get } 1=\tan 45^{\circ}=\frac{2 \sqrt{h^2-a b}}{a+b} }

\mathrm{ (a+b)^2=4\left(h^2-a b\right) }

\mathrm{ \Rightarrow\left(4 a^2+4 a \alpha\right)^2=4\left[4 a^2 \beta^2-4\left(a^2\right) 4 a \alpha\right] }

\mathrm{ \Rightarrow(a+\alpha)^2=\beta^2-4 a \alpha }

\mathrm{ \Rightarrow \alpha^2+6 a \alpha+a^2-\beta^2=0 \Rightarrow(\alpha+3 a)^2-\beta^2=8 a^2 }

Thus, the equation of the required locus is \mathrm{ (x+3 a)^2-y^2=8 a^2.}

Hence option 4 is correct.




 


 

Posted by

Gaurav

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