Question

The equation of the tangent to the parabola ${{y}^{2}}=4x$ is given by $3x+4y=4$. Find the contact points?

Answer 1

In this type of question we have to use the concept of finding contact points. We know that if we have to find the contact points of the equation of tangent to the parabola then it is equivalent to find the point of intersection. Here we express y in terms of x by using the equation of any of the curves and then we substitute this expression of y in the equation of another curve and on simplification we can obtain the required result.

Complete step-by-step solution:
Now we have to find the contact points for the parabola ${{y}^{2}}=4x$ and its tangent whose equation is given by $3x+4y=4$

Now we can rewrite the equation of tangent as

& \Rightarrow 3x+4y=4 \\\ & \Rightarrow 4y=4-3x \\\ & \Rightarrow y=\dfrac{4-3x}{4}\cdots \cdots \cdots \left( i \right) \\\ \end{aligned}$$ By substituting this value of $$y$$ in the equation of parabola we can write, $$\begin{aligned} & \Rightarrow {{y}^{2}}=4x \\\ & \Rightarrow {{\left( \dfrac{4-3x}{4} \right)}^{2}}=4x \\\ & \Rightarrow \dfrac{16-24x+9{{x}^{2}}}{16}=4x \\\ & \Rightarrow 16-24x+9{{x}^{2}}=64x \\\ & \Rightarrow 9{{x}^{2}}-24x-64x+16=0 \\\ & \Rightarrow 9{{x}^{2}}-88x+16=0 \\\ \end{aligned}$$ As we know that the solution of the quadratic equation $$a{{x}^{2}}+bx+c=0$$ is given by $$\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$$, hence the solution of the above quadratic equation is $$\begin{aligned} & \Rightarrow x=\dfrac{-\left( -88 \right)\pm \sqrt{{{\left( -88 \right)}^{2}}-4\left( 9 \right)\left( 16 \right)}}{2\times 9} \\\ & \Rightarrow x=\dfrac{88\pm \sqrt{7744-576}}{18} \\\ & \Rightarrow x=\dfrac{88\pm \sqrt{7168}}{18} \\\ & \Rightarrow x=\dfrac{88\pm 84.66}{18} \\\ & \Rightarrow x=\dfrac{88+84.66}{18}\text{ and }x=\dfrac{88-84.66}{18} \\\ & \Rightarrow x=9.6\text{ and }x=0.2 \\\ \end{aligned}$$ Hence, we get the values of $$x$$ Now, by substituting these values of $$x$$ in equation $$\left( i \right)$$, we can obtain the values for $$y$$ $$\Rightarrow y=\dfrac{4-3x}{4}=\dfrac{4-\left( 3\times 9.6 \right)}{4}=-6.2\text{ and }y=\dfrac{4-3x}{4}=\dfrac{4-\left( 3\times 0.2 \right)}{4}=0.9$$ Thus we get the points of intersection as $$\Rightarrow \left( 9.6,-6.2 \right)\text{ and }\left( 0.2,0.9 \right)$$ Hence, the contact points for the parabola $${{y}^{2}}=4x$$ and its tangent whose equation is given by $$3x+4y=4$$ are $$\left( 9.6,-6.2 \right)\text{ and }\left( 0.2,0.9 \right)$$ **Note:** In this type of question students have to remember the steps to find the contact points i.e. the points of intersection of any two intersecting curves. Students also have to remember the formula for calculation of roots of quadratic equations. Also students have to note that as we have to find the contact points the final answer must be in the form of $$\left( a,b \right)$$.