Question

Show that the line \[x + 2y + 8 = 0\] is tangent to the parabola \[{y^2} = 8x\]. Hence, find the point of contact.

Answer 1

Hint: Determine the equation of the tangent to the parabola \[{y^2} = 8x\] and show that the line \[x + 2y + 8 = 0\] satisfies the equation and then find the point of contact.

Complete step by step answer:
The equation of the given parabola is \[{y^2} = 8x\].

We can find the slope of the tangent to the parabola by differentiating the equation of the parabola.

\[{y^2} = 8x.........(1)\]

Differentiating both sides of equation (1) with respect to x, we get:

\[2y\dfrac{{dy}}{{dx}} = 8\]

Solving for \[\dfrac{{dy}}{{dx}}\] from the above equation, we get:

\[\dfrac{{dy}}{{dx}} = \dfrac{8}{{2y}}\]

We can cancel 8 and 2 to get 4 in the numerator.

\[\dfrac{{dy}}{{dx}} = \dfrac{4}{y}..........(2)\]

If the tangent’s point of contact with the parabola is (h, k), then the slope m of this tangent using equation (2) is given by:

\[m = \dfrac{{dy}}{{dx}} = \dfrac{4}{k}...........(3)\]

We can find the equation of a line using slope and a point it passes through using the slope-point formula. The equation of the line passing through a point (a, b) and having slope m is given as follows:

\[y - b = m(x - a)..........(4)\]

Substituting equation (3) in (4), the equation of the line passing through the point (h, k) is given as follows:


\[y - k = \dfrac{4}{k}(x - h)\]

Taking k to the other side and multiplying, we get:

\[k(y - k) = 4(x - h)\]

\[ky - {k^2} = 4x - 4h\]

Writing in standard form, we get:

\[4x - ky + {k^2} - 4h = 0............(5)\]

For two lines to be equal, the coefficients in the equations must be proportionate.

Comparing equation (5) with the line \[x + 2y + 8 = 0\], we have:

\[\dfrac{4}{1} = \dfrac{{ - k}}{2} = \dfrac{{{k^2} - 4h}}{8}\]

Taking the first two equal terms and solving, we get the value of k.

\[\dfrac{4}{1} = \dfrac{{ - k}}{2}\]

\[k = - 8\]

Now considering the first and the third equal terms, and substituting the value of k, we have:

\[\dfrac{4}{1} = \dfrac{{{{( - 8)}^2} - 4h}}{8}\]

\[32 = 64 - 4h\]

Solving for h, we have:

\[ - 4h = - 32\]

\[h = \dfrac{{ - 32}}{{ - 4}}\]

\[h = 8\]

We get a real value for h and k, implying that the given line \[x + 2y + 8 = 0\] is the equation of the tangent to the parabola \[{y^2} = 8x\].

The point of contact is (8, - 8).

Note: You can also directly use the formula for the tangent to a parabola \[{y^2} = 4ax\], that is, \[y = mx + \dfrac{a}{m}\] with point of contact \[\left( {\dfrac{a}{{{m^2}}},\dfrac{{2a}}{m}} \right)\].