Using Discriminant to find equation of a line?

Question

Find the equation of the tangent to the parabola $$y = x^2 − 5x − 3$$ that is parallel to the line $3x − y − 7 = 0$.

I know how to solve this question utilizing differentiation, but I can't think of a way to solve it using quadratic discriminant theory. Any help would be greatly appreciated.

Answer 1

The second equation in the question can be rearranged to:
$$ y = 3x - 7 $$
Now the $-7$ is irrelevant, as we are only interested in the slope which is $3$.
So let $$ y = 3x + c \tag{1}$$ be the form of the answer that we want. We need to find a value of $c$ that is a tangent. Now solve this equation and the quadratic given in the question as simultaneous equations: $$ x^2 -5x -3 = 3x +c $$ When simplified and arranged in standard form is: $$ x^2 -8x - (c+3) = 0 $$ Now letting the discriminant of the above equation equal $0$, because a tangent has only one intersection to the parabola, we have: $$ b^2 - 4ac = 0 $$ Substitute the values into the discriminant equation we get: $$ (-8)^2 -4 \times1 \times-(c+3) =0 $$ Solving this equation we find that: $$ c = -19 $$ Now substitute this value for $c$ into the original straight line that we took from the question, I have labeled it ($1$) above. We get: $$ y = 3x -19 $$ Or rearranging to standard from: $$ 3x - y -19 = 0 $$

Answer 2

Hint:

You want to solve an equation which has a repeated root (that has something to do with the discriminant, right?). This equation is the intersection of the parabola and the line that has slop $3$.

Answer 3

Given line $3x-y-7=0$

Required line is parallel to this line So we can take that as $3x-y+k=0$----->(1)

Given parabola $y=x^2-5x-3$---->(2)

Solving (1)and(2)

$3x+k=x^2-5x-3$

$\implies x^2-8x-3-k=0$

Which is a quadratic equation in terms of x

If it's a tangent then it's discriminent is zero

$\Delta=0$

$(-8)^2-4(1)(-3-k)=0$

$implies 4k=-76$

$\implies k=-19$

Now required tangent is

$ 3x-y-19=0$