Question:
Consider the parabola with vertex \(\left(\frac{1}{2},\frac{3}{4}\right)\) and the directrix \(y=\frac{1}{2}\). Let P be the point where the parabola meets the line \(x=-\frac{1}{2}\). If the normal to the parabola at P intersects the parabola again at the point Q, then \((PQ)^2\) is equal to :
Consider the parabola with vertex \(\left(\frac{1}{2},\frac{3}{4}\right)\) and the directrix \(y=\frac{1}{2}\). Let P be the point where the parabola meets the line \(x=-\frac{1}{2}\). If the normal to the parabola at P intersects the parabola again at the point Q, then \((PQ)^2\) is equal to :
Show Hint
The focus is \(\left(\frac12,1\right)\), so \(a=\frac14\) and the parabola is \(\left(x-\frac12\right)^2=y-\frac34\). Find P, the normal at P, and its second intersection Q.
Updated On: Nov 4, 2026
- \(\frac{75}{8}\)
- \(\frac{125}{16}\)
- \(\frac{25}{2}\)
- \(\frac{15}{2}\)
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The Correct Option is B
Solution and Explanation
Step 1: Vertex \(\left(\frac12,\frac34\right)\), directrix \(y=\frac12\) is below the vertex at distance \(\frac14\), so \(a=\frac14\) and the parabola opens upward: \(\left(x-\frac12\right)^2=4a\left(y-\frac34\right)=y-\frac34\).
Step 2: At \(x=-\frac12\): \(1=y-\frac34\Rightarrow y=\frac74\). So \(P\left(-\frac12,\frac74\right)\).
Step 3: \(y=\frac34+\left(x-\frac12\right)^2\Rightarrow \frac{dy}{dx}=2\left(x-\frac12\right)=-2\) at P. Slope of normal = \(\frac12\).
Step 4: Normal: \(y-\frac74=\frac12\left(x+\frac12\right)\). Put \(u=x-\frac12\): \(y=\frac74+\frac{u+1}{2}\). Equate with \(y=\frac34+u^2\): \(u^2-\frac u2-\frac32=0\Rightarrow 2u^2-u-3=0\Rightarrow u=\frac32\) or \(u=-1\).
Step 5: \(u=-1\) is P, so for Q, \(u=\frac32\): \(x=2,\ y=\frac34+\frac94=3\). \(Q(2,3)\).
Step 6: \(PQ^2=\left(\frac52\right)^2+\left(3-\frac74\right)^2=\frac{25}{4}+\frac{25}{16}=\frac{125}{16}\).
Answer: (2) \(\frac{125}{16}\)
Step 2: At \(x=-\frac12\): \(1=y-\frac34\Rightarrow y=\frac74\). So \(P\left(-\frac12,\frac74\right)\).
Step 3: \(y=\frac34+\left(x-\frac12\right)^2\Rightarrow \frac{dy}{dx}=2\left(x-\frac12\right)=-2\) at P. Slope of normal = \(\frac12\).
Step 4: Normal: \(y-\frac74=\frac12\left(x+\frac12\right)\). Put \(u=x-\frac12\): \(y=\frac74+\frac{u+1}{2}\). Equate with \(y=\frac34+u^2\): \(u^2-\frac u2-\frac32=0\Rightarrow 2u^2-u-3=0\Rightarrow u=\frac32\) or \(u=-1\).
Step 5: \(u=-1\) is P, so for Q, \(u=\frac32\): \(x=2,\ y=\frac34+\frac94=3\). \(Q(2,3)\).
Step 6: \(PQ^2=\left(\frac52\right)^2+\left(3-\frac74\right)^2=\frac{25}{4}+\frac{25}{16}=\frac{125}{16}\).
Answer: (2) \(\frac{125}{16}\)
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