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  • Find the value of  if the equation <img alt="(x+1)^{2}+(y-1)^{2}=\lambda(2x+y+6)^{2}" src="https://lear

Find the value of \lambda if the equation (x+1)^{2}+(y-1)^{2}=\lambda(2x+y+6)^{2} represents a parabola. Also, find its focus.

Option: 1

\lambda=\frac{1}{2} \;\;\; \&\;\;\; \text { Its focus is }(1,-1)


Option: 2

\lambda=\frac{1}{5} \;\;\; \&\;\;\; \text { Its focus is }(1,-1)


Option: 3

\lambda=\frac{1}{2} \;\;\; \&\;\;\; \text { Its focus is }(-1,1)


Option: 4

\lambda=\frac{1}{5} \;\;\; \&\;\;\; \text { Its focus is }(-1,1)


Answers (1)

best_answer

 

 

Equation of parabola when equation of axis, tangent at vertex and latusrectum are given -

Equation of parabola when equation of axis, tangent at vertex and latusrectum are given

Let the equation of axis be lx + my + n = 0 is given then the equation of the tangent at the vertex will be mx - ly + r = 0.

 

Equation of parabola is

\\\mathrm{\;\;\;\;\;\;\;\;\;(PM)^2=(Latusrectum)\cdot (PN)}\\\\\mathrm{\Rightarrow \left ( \frac{lx+my+n}{\sqrt{l^2+m^2}} \right )^2=(Latusrectum)\cdot\left ( \frac{mx-ly+r}{\sqrt{m^2+n^2}} \right )}

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\begin{array}{l}{(x+1)^{2}+(y-1)^{2}=\lambda(2x+y+6)^{2}=5 \lambda\left(\frac{2x+y+6}{\sqrt{5}}\right)^{2}} \\ {\therefore \quad \sqrt{(x+1)^{2}+(y-1)^{2}}=\sqrt{5 \lambda} \cdot \frac{|2x+y+6|}{\sqrt{5}}} \\\\ {\text { This represents parabola if } \sqrt{5 \lambda}=1} \\ {\Rightarrow \quad \lambda=\frac{1}{5}} \\\\ {\text { Its focus is }(-1,1)}\end{array}

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