Question

The equation of common tangent to the parabola, \[{{y}^{2}}=2x\] and \[{{x}^{2}}=16y\] is:
\[Ax+By+C=0\], where A, B, C are integers then the values of A, B, C
(a) 1, 2, 2
(b) 2, 2, 1
(c) 2, 1, 2
(d) none of these

Answer 1

Hint: In this question we first need to find the points of intersections of the given two parabolas. Now, the common tangent should pass through those points of intersection which can be found using the two point formula which is given by \[\dfrac{x-{{x}_{1}}}{{{x}_{2}}-{{x}_{1}}}=\dfrac{y-{{y}_{1}}}{{{y}_{2}}-{{y}_{1}}}\]. Now, further simplification gives the equation which in comparison with the given equation in the question gives the values of A, B, C.

Complete step by step solution:
Now, the equations of tangents given in the question are
\[{{y}^{2}}=2x,{{x}^{2}}=16y\]
Now, let us solve these two equations to get the points of intersection
Now, from the given first parabola we get,
\[\Rightarrow x=\dfrac{{{y}^{2}}}{2}\]
Now, let us substitute this value of x in the given second equation of parabola
\[\Rightarrow {{x}^{2}}=16y\]
Now, on substituting the respective value we get,
\[\Rightarrow {{\left( \dfrac{{{y}^{2}}}{2} \right)}^{2}}=16y\]
Now, this can be further written as
\[\Rightarrow \dfrac{{{y}^{4}}}{4}=16y\]
Now, on cancelling out the common terms and rearranging them we get,
\[y=0\]
\[\Rightarrow {{y}^{3}}=64\]
Now, on applying the cube root on both sides we get,
\[\Rightarrow {{\left( {{y}^{3}} \right)}^{\dfrac{1}{3}}}={{\left( 64 \right)}^{\dfrac{1}{3}}}\]
Now, on further simplification we get,
\[\therefore y=4,y=0\]
Now, on substituting this value of y in the second parabola we get,
\[\Rightarrow {{x}^{2}}=16\times 4\]
Now, on applying the square root on both sides we get,
\[\Rightarrow {{\left( {{x}^{2}} \right)}^{\dfrac{1}{2}}}={{\left( 64 \right)}^{\dfrac{1}{2}}}\]
Now, on further simplification we get,
\[\therefore x=8,x=0\]
Now, the point of intersections of the two parabolas are
\[\left( 8,4 \right),\left( 0,0 \right)\]

Now, the common tangent of these two parabolas passes through these two points
Now, on comparing with the two point formula we have
\[{{x}_{1}}=8,{{y}_{1}}=4,{{x}_{2}}=0,{{y}_{2}}=0\]
Now, from the two point formula we have
\[\Rightarrow \dfrac{x-{{x}_{1}}}{{{x}_{2}}-{{x}_{1}}}=\dfrac{y-{{y}_{1}}}{{{y}_{2}}-{{y}_{1}}}\]
Now, on substituting the respective values we get,
\[\Rightarrow \dfrac{x-8}{0-8}=\dfrac{y-4}{0-4}\]
Now, on cancelling the common terms and rearranging we get,
\[\Rightarrow x-8=2\left( y-4 \right)\]
Now, on further rearranging the terms we get,
\[\Rightarrow x-2y=0\]
Now, on comparing with the given equation \[Ax+By+C=0\] we get,
\[A=1,B=-2,C=0\]
Hence, the correct option is (d).

Note: Instead of finding the points of intersection and simplifying further we can also solve this by considering the equation of tangent for first parabola and then equating it to the equation of tangent of the second parabola which also gives the same result.
It is important to note that while substituting the values and simplifying further we need to substitute the respective values because changing any one of them changes the result.