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  • If the tangent at a point on the parabola  <img alt="y^2=3 x" src="https://entrancecorner.oncodecogs.com/gif.latex?y%5E2

If the tangent at a point P on the parabola  y^2=3 x is parallel to the line x+2 y=1 and the tangents at the points Q and R on the ellipse \frac{x^2}{4}+\frac{y^2}{1}=1 are perpendicular to the line x-y=2, then the area of the triangle \mathrm{PQR} is :

Option: 1

\frac{3}{2} \sqrt{5}


Option: 2

3 \sqrt{5}


Option: 3

\frac{9}{\sqrt{5}}


Option: 4

5 \sqrt{3}


Answers (1)

best_answer

\begin{aligned} & x+2 y=1 \, \, \, \, \, \, \, \, \, \quad y^2=3 x \\ & \mathrm{~m}=-\frac{1}{2}\, \, \, \, \, \, \, \, \quad \mathrm{~T}_{\mathrm{p}}: y=-\frac{1}{2} \mathrm{x}+\frac{\frac{3}{4}}{-\frac{1}{2}} \end{aligned}

                                         \begin{aligned} & y=-\frac{x}{2}-\frac{3}{2} \\ & 2 y+x+3=0\, \, \, \, \, \, \, \, \, ....(1) \end{aligned}

\begin{array}{ll} x-y=2 & \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, E: \frac{x^2}{4}+\frac{y^2}{1}=1 \\ m=1 \, \, \, \, & \, \, \, \, \, \, \, \, \, \, \, \, \, \, \,\, \, \, \, \, \, \, y=-x \pm \sqrt{(-1)^2 4+1} \end{array}

slope of tangent at Q & R is –1      \begin{aligned} &\begin{aligned} & y=-x \pm \sqrt{5} \\ & x+y=\sqrt{5} \end{aligned}\\ &\text {.... (2) } x+y=-\sqrt{5}\, \, \, ...(3) \end{aligned}

  Point P :                                       Point Q :                                                   Point R:   \mathrm{T}=\mathrm{O} \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \quad \frac{\mathrm{xx}_2}{4}+\frac{\mathrm{yy}_2}{1}=1 \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \quad \frac{\mathrm{x}_2}{1}=\frac{4 \mathrm{y}_2}{1}=\frac{-4}{\sqrt{5}}\mathrm{yy}_1=\frac{3}{2}\left(x+x_1\right) \, \, \, \, \, \, \, \, \, \quad \, \, xx_2+4 \mathrm{yy}_2-4=0 \, \, \, \, \, \, \, \, \, \, \ \quad x_2=\frac{-4}{\sqrt{5}}, \mathrm{y}=\frac{1}{\sqrt{5}}

3 x-2 y_1+3 x_1=0 \, \, \, \, \, \, \, \, \, \, \quad \frac{x_2}{1}=\frac{4 y_2}{1}=\frac{-4}{-\sqrt{5}}

Comparison with (1)                     x_2=\frac{4}{\sqrt{5}}-y_2=\frac{1}{\sqrt{5}}                                                                

\begin{aligned} & \frac{3}{1}=\frac{-2 \mathrm{y}_1}{2}=\frac{3 \mathrm{x}_1}{3} \\ & \mathrm{y}_1=-3, \quad \mathrm{x}_1=3 \end{aligned}

\text { Area of } \triangle \mathrm{PQR}

\begin{aligned} & =\frac{1}{2}\left|\begin{array}{ccc} 3 & -3 & 1 \\ \frac{4}{\sqrt{5}} & \frac{1}{\sqrt{5}} & 1 \\ -\frac{4}{\sqrt{5}} & -\frac{1}{\sqrt{5}} & 1 \end{array}\right| \\ & \Rightarrow \frac{1}{2}\left[3\left(\frac{1}{\sqrt{5}}+\frac{1}{\sqrt{5}}\right)+3\left(\frac{4}{\sqrt{5}}+\frac{4}{\sqrt{5}}\right)+1\left(-\frac{4}{5}+\frac{4}{5}\right)\right] \\ & \Rightarrow \frac{1}{2}\left[\frac{6}{\sqrt{5}}+\frac{24}{\sqrt{5}}\right] \\ & \Rightarrow \frac{1}{2} \times \frac{30}{\sqrt{5}}=\frac{5 \times 3}{\sqrt{5}}=3 \sqrt{5} \\ & \end{aligned}

 

Posted by

Divya Prakash Singh

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