Prove the straight line as a tangent to a parabola.

Question

I was going through some past exam papers and I came across this problem and I'm bit puzzled on how to approach this, could someone please help me out?

Equation of the parabola $y^2-7x-8y+14=0$, prove that the straight line given by $7x+6y=13$ is a tangent and find the point of contact.

Answer 1

given parabola : $y^2-7x-8y+14=0$,

given line : $7x+6y=13$

point of contact : $y^2-7x-8y+14=y^2+(6y-13)-8y+14=0$ i.e., $y^2-2y+1=0$

i.e., $y=1$ i.e., $(1,1)$ is "the" point of contact

when "a" point of contact is "the" point of contact.. It must be tangent...

Answer 2

Hint: First find the derivative using implicit differentiation.

Then you want to find the point where the slope of the tangent is $-7/6$. That is, you ask the question: When is the derivative equal to $-7/6$? From what I can see, this equation will give you the $y$-coordinate to the point. Now take that and find the $x$-coordinate from the original equation.