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Question: If the line \(x-1=0\) is the directrix of the parabola \({{y}^{2}}-kx+8=0\), then find the value of ...

If the line x−1=0x-1=0 is the directrix of the parabola y2−kx+8=0{{y}^{2}}-kx+8=0, then find the value of k.
A. 18\dfrac{1}{8}
B. 8
C. 4
D. 14\dfrac{1}{4}

Explanation

Solution

We find the general equation of the parabola. We find the focus, vertex, and directrix equation. Then we equate it with the given equation. We find three equations of three unknowns. We solve them to get a quadratic equation of k. By solving that we find the possible value of k.

Complete step-by-step answer:
The general equation of a parabola is (y−n)2=4a(x−m){{\left( y-n \right)}^{2}}=4a\left( x-m \right). Its vertex is (m,n)\left( m,n \right). Focus is (m+a,n)\left( m+a,n \right). The equation of the directrix will be x=m−ax=m-a.
We have been given that the line x−1=0x-1=0 is the directrix of the parabola y2−kx+8=0{{y}^{2}}-kx+8=0.
We make a comparison of the given equation with the general equation of parabola.
y2−kx+8=0 ⇒y2=kx−8 ⇒y2=k(x−8k) \begin{aligned} & {{y}^{2}}-kx+8=0 \\\ & \Rightarrow {{y}^{2}}=kx-8 \\\ & \Rightarrow {{y}^{2}}=k\left( x-\dfrac{8}{k} \right) \\\ \end{aligned}
Equating with (y−n)2=4a(x−m){{\left( y-n \right)}^{2}}=4a\left( x-m \right) we get n=0,4a=k,m=8kn=0,4a=k,m=\dfrac{8}{k}.
Also, we have directrix as x−1=0x-1=0. Equating with x=m−ax=m-a, we get 1=m−a1=m-a.
We need to find the possible values of k.
We have k=4ak=4a …(i), m=8km=\dfrac{8}{k} …(ii), m−a=1m-a=1 …(iii)
We find value of a and m from equation (i), (ii) and put it in equation (iii)
So, k=4a⇒a=k4k=4a\Rightarrow a=\dfrac{k}{4} and m=8km=\dfrac{8}{k}
Putting in the equation (iii) we get
m−a=1 ⇒8k−k4=1 \begin{aligned} & m-a=1 \\\ & \Rightarrow \dfrac{8}{k}-\dfrac{k}{4}=1 \\\ \end{aligned}
We got a quadratic equation of k as
8k−k4=1 ⇒32−k24k=1 ⇒32−k2=4k ⇒k2+4k−32=0 \begin{aligned} & \dfrac{8}{k}-\dfrac{k}{4}=1 \\\ & \Rightarrow \dfrac{32-{{k}^{2}}}{4k}=1 \\\ & \Rightarrow 32-{{k}^{2}}=4k \\\ & \Rightarrow {{k}^{2}}+4k-32=0 \\\ \end{aligned}
Now we solve the equation to find value of k.
k2+4k−32=0 ⇒k2+8k−4k−32=0 ⇒(k+8)(k−4)=0 \begin{aligned} & {{k}^{2}}+4k-32=0 \\\ & \Rightarrow {{k}^{2}}+8k-4k-32=0 \\\ & \Rightarrow \left( k+8 \right)\left( k-4 \right)=0 \\\ \end{aligned}
From the equation we find k=4,−8k=4,-8. Both are correct. We need to match it with given options.

So, the correct answer is “Option C”.

Note: We need to cross-check the value of k from the given equation. We need to confirm if that value satisfies all the conditions. Instead of taking general equations we can take the efficient one where we already have an idea about the vertex points.