\(y^2=6x\; \&\;y^2=4ax\)
\(⇒4a=6⇒a=23\)
\(y=mx+2m^3;(m\neq0)\)
\(h=\frac{6m−3}{4m^2},k=\frac{6m+3}{4m}\), Now eliminating m and we get
\(⇒3h=2(−2k^2+9k−9)\)
\(⇒4y^2−18y+3x+18=0\)
\(\text{The Correct Option is (C):}\) \(4 y^2+18 y+3 x+18=0\)
Let P be a point on the parabola y2 = 4ax, where a > 0. The normal to the parabola at P meets the x -axis at a point Q. The area of the triangle PFQ where F is the focus of the parabola, is 120. If the slope m of the normal and a are both positive integers, then the pair (a, m) is
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