Question

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) $p=2$, $h=3$, $k=-4$.

Answer 1

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.