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# If a chord, which is not a tangent, of the parabola ${{y}^{2}}=16x$ has the equation $2x+y=p$, and midpoint $\left( h,k \right)$, then which of the following is (are) possible value(s) of p, h and k?(a) $p=-1$, $h=1$, $k=-3$.(b) $p=5$, $h=4$, $k=-3$,(c) $p=-2$, $h=2$, $k=-4$,(d) $p=2$, $h=3$, $k=-4$.

Last updated date: 10th Sep 2024
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Hint: We start solving the problem by finding the equation of the chord of the parabola ${{y}^{2}}=16x$ having midpoint $\left( h,k \right)$. We then equate the obtained equation of chord with the given equation of chord $2x+y=p$. We then take the ratios of coefficients of x and y and the constants of respective lines will be equal to each other. We then solve these obtained ratios to get the required answer.

Complete step-by-step solution:
Let us first draw the given information:

Consider the end points of chord having midpoint at $\left( h,k \right)$of the parabola ${{y}^{2}}=16x$ or ${{y}^{2}}=4\times 4\times x$ are $\left( {{x}_{1}},{{y}_{1}} \right)$ and $\left( {{x}_{2}},{{y}_{2}} \right)$. We know the section formula the midpoint of the line joining $\left( {{x}_{1}},{{y}_{1}} \right)$ and $\left( {{x}_{2}},{{y}_{2}} \right)$ is given by $\left( \dfrac{{{x}_{1}}+{{x}_{2}}}{2},\dfrac{{{y}_{1}}+{{y}_{2}}}{2} \right)$. Now we see that the coordinates $\left( \dfrac{{{x}_{1}}+{{x}_{2}}}{2},\dfrac{{{y}_{1}}+{{y}_{2}}}{2} \right)$ and $\left( h,k \right)$ are identical. Therefore, equating the ordinates we will get
$\Rightarrow k=\dfrac{{{y}_{1}}+{{y}_{2}}}{2}$.
$\Rightarrow 2k={{y}_{2}}+{{y}_{1}}$--- (1).
As we know that the points $\left( {{x}_{1}},{{y}_{1}} \right)$ and $\left( {{x}_{2}},{{y}_{2}} \right)$ must lie on the parabola ${{y}^{2}}=16x$ then the coordinates must satisfy the equation of parabola, hence we obtain,
${{y}_{1}}^{2}=16{{x}_{1}}$ ---(2) and ${{y}_{2}}^{2}=16{{x}_{2}}$ ---(3).
Subtracting Eq. (2) from eq. (3) we will get,
$\Rightarrow {{y}_{2}}^{2}-{{y}_{1}}^{2}=16\left( {{x}_{2}}-{{x}_{1}} \right)$.
$\Rightarrow \left( {{y}_{2}}+{{y}_{1}} \right)\left( {{y}_{2}}-{{y}_{1}} \right)=16\left( {{x}_{2}}-{{x}_{1}} \right)$ --- (4).
Now substituting the value of eq. (1) in eq. (4), we will get
$\Rightarrow 2k\left( {{y}_{2}}-{{y}_{1}} \right)=16\left( {{x}_{2}}-{{x}_{1}} \right)$.
$\Rightarrow \dfrac{\left( {{y}_{2}}-{{y}_{1}} \right)}{\left( {{x}_{2}}-{{x}_{1}} \right)}=\dfrac{8}{k}$ ---(5).
Here we got the slope of the chord joining $\left( {{x}_{1}},{{y}_{1}} \right)$and. is$\dfrac{8}{k}$. Now the equation of the line (the chord) passing through the point $\left( h,k \right)$ and having slope $\dfrac{8}{k}$is given by,
$\Rightarrow \left( y-k \right)=\dfrac{8}{k}\left( x-h \right)$.
$\Rightarrow 8x-ky=8h-{{k}^{2}}$ ---(6).
But we are given that the equation of the chord is $2x+y=p$---(7).
Since line eq. (6) and eq. (7) are identical, then the ratio of the respective coefficients must be equal therefore we will get
$\Rightarrow \dfrac{8}{2}=\dfrac{-k}{1}=\dfrac{8h-{{k}^{2}}}{p}$.
$\Rightarrow 4=-k=\dfrac{8h-{{k}^{2}}}{p}$.
$\Rightarrow k=-4$, $8h-{{k}^{2}}=4p$.
$\Rightarrow k=-4$, $8h-16=4p$---(8).
We can see that the options (c) and (d) have $k=-4$. So, let us substitute the values of h and k in the equation to get the correct options.
Let us substitute option (c) $p=-2$, $h=2$, $k=-4$ in equation (8) to verify them.
So, we get $8\left( 2 \right)-16=4\left( -2 \right)$.
$\Rightarrow 0=-8$, which is a contradiction.
So, option (c) is not correct.
Let us substitute option (d) $p=2$, $h=3$, $k=-4$ in equation (8) to verify them.
So, we get $8\left( 3 \right)-16=4\left( 2 \right)$.
$\Rightarrow 8=8$, which is true.
So, option (d) is correct.
$\therefore$ The correct option for the given problem is (d).

Note: Alternatively, we can solve the problem by using the formula of the chord of the parabola ${{y}^{2}}=4ax$ having midpoint $\left( {{x}_{1}},{{y}_{1}} \right)$ is $y{{y}_{1}}-4ax={{y}_{1}}^{2}-4a{{x}_{1}}$. We should not make calculation mistakes while solving this problem. We can also use the parametric points of the parabola $\left( a{{t}^{2}},2at \right)$ to solve this problem, but this increases one more variable which may lead us to confusion.