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  • If y=mx+4 is a tangent to both the parabolas,  and  

If y=mx+4 is a tangent to both the parabolas, y^{2}=4x and x^{2}=2by, then b is equal to :
Option: 1 -64
Option: 2 128
Option: 3 -128
Option: 4 -64
 

Answers (1)

best_answer

 

 

Tangents of Parabola in Slope Form -

Tangents of Parabola in Slope Form

\\ {\text { Equation of the tangent to the parabola } \mathrm{y}^{2}=4 \mathrm{ax} \text { at the point } \mathrm{P}\left(\mathrm{x}_{1}, \mathrm{y}_{1}\right) \text { is }} \\ {\mathrm{y} \mathrm{y}_{1}=2 \mathrm{a}\left(\mathrm{x}+\mathrm{x}_{1}\right)}

m is the slope of the tangent, then

\\ {\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\mathrm{m}=\frac{2 \mathrm{a}}{\mathrm{y}_{1}} \Rightarrow \mathrm{y}_{1}=\frac{2 \mathrm{a}}{\mathrm{m}}} \\ {\left(\mathrm{x}_{1}, \mathrm{y}_{1}\right) \text { lies on the paarabola } \mathrm{y}^{2}=4 \mathrm{ax}} \\ {\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\mathrm{y}_{1}^{2}=4 \mathrm{ax}_{1} \Rightarrow \frac{4 \mathrm{a}^{2}}{\mathrm{m}^{2}}=4 \mathrm{ax}_{1}} \\ {\therefore\;\;\;\;\;\;\;\;\;\;\; \mathrm{x}_{1}=\frac{\mathrm{a}}{\mathrm{m}^{2}}} \\ {\text { put the value of } \mathrm{x}_{1} \text { and } \mathrm{y}_{1} \text { in the equation } \mathrm{yy}_{1}=2 \mathrm{a}\left(\mathrm{x}+\mathrm{x}_{1}\right)} \\ {\text { we get }} \\ {\Rightarrow \;\;\;\;\;\;\;\;\;\;\mathrm{y}=\mathrm{mx}+\frac{\mathrm{a}}{\mathrm{m}}} \\ {\text { which is equation of tangent of the parabola in slope form. }}
The coordinates of point of contact are 

\left(\frac{a}{m^{2}}, \frac{2 a}{m}\right)

TIP

\begin{array}{c||cc} \mathbf { Equation \;of \;Parabola } & {\mathbf { Point \;of \;Contact }} & {\mathbf { Tangent\; Equation }} \\\\ \hline\hline \\ {\color{Black} y^{2}{=4 a x}} & {\color{Black} {\left(\frac{a}{m^{2}}, \frac{2 a}{m}\right)}} & {\color{Black} {y=m x+\frac{a}{m}}}\\ \\ {\color{Black} y^{2}{=-4 a x}} & {\color{Black} {\left(-\frac{a}{m^{2}}, \frac{2 a}{m}\right)}} & {\color{Black} {y=m x-\frac{a}{m}}} \\\\ {\color{Black} x^{2}{=4 a y}} & {\color{Black} {\left(2am, am^2\right)}} & {\color{Black} {y=m x-am^2}} \\\\ {\color{Black} x^{2}{=-4 a y}} & {\color{Black} {\left(2am, -am^2\right)}} & {\color{Black} {y=m x+am^2}} \end{array}
 

-

\begin{align*} y&= mx-am^2 \\ y&=\frac{x}{4}+4\\ y=4x , y& = mx+\frac{1}{m}\\ \Rightarrow m&=\frac{1}{4} \\ y &= \frac{x}{4}+4 \end{align*}

also ,

 x^2=2by \\ y = mx -\frac{bm^2}{2}\\ y= \frac{x}{4}+4\\ \implies \frac{bm^2}{2}=-4\\

since , m= \frac{1}{4}

Therefore , b = \textbf{-128}

Correct Option (3)

Posted by

Ritika Jonwal

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