### Video Transcript

A tangent to π₯ squared plus π¦
squared equals 72 forms an isosceles triangle when taken with the positive π₯- and
π¦-axes. What is the equation of this
tangent?

Letβs begin by sketching the curve
and the tangent in question. π₯ squared plus π¦ squared equals
72 is the equation of a circle with its center at the origin and a radius of root 72
units. There is then a tangent to this
circle which forms an isosceles triangle with the positive π₯- and π¦-axes. We know that it is the two sides
that lie along the π₯- and π¦-axes that must be the two equal sides of this
isosceles triangle. Because this is also a right
triangle and in any right triangle, the hypotenuse is the longest side and is
strictly longer than any other side in the triangle, the two shorter sides must
therefore be the sides of equal length.

In order to find the equation of a
tangent or indeed any straight line, we need to know its slope and the coordinates
of at least one point that lies on the line. Letβs begin by considering the
slope of this tangent. The slope of the tangent will be
the same as the slope of the curve at the point of contact. We donβt know the coordinates of
this point, but letβs begin by finding the slope function of the curve. Weβll do this using
differentiation. As the curve is defined implicitly,
weβll need to use implicit differentiation when differentiating the second term with
respect to π₯. Differentiating each term with
respect to π₯ then gives two π₯ plus two π¦ dπ¦ by dπ₯ equals zero.

We can make dπ¦ by dπ₯ the subject
of this equation by first subtracting two π₯ from each side and then dividing by two
π¦, giving dπ¦ by dπ₯ equals negative two π₯ over two π¦. Finally, canceling a factor of two
gives dπ¦ by dπ₯ equals negative π₯ over π¦.

So weβve found an expression for
the slope function of the curve, but we donβt have the coordinates of the point
where the tangent meets the curve to substitute in. What we do know is that the
triangle formed by the tangent and the positive π₯- and π¦-axes is isosceles. So if the coordinates of the point
where the tangent intersects the π₯-axis are π, zero, then the coordinates of the
point where the tangent intersects the π¦-axis will be zero, π so that the two line
segments in pink are of equal length.

We can then recall that another way
to find the slope of a straight line given two points that lie on the line is using
the formula π equals π¦ two minus π¦ one over π₯ two minus π₯ one. Substituting the two points with
coordinates zero, π and π, zero and then simplifying gives π equals negative
one.

So we now know that the slope of
the tangent is negative one. In order to find its equation, we
need to determine the coordinates of the point that lies on the tangent. At the point where the tangent
meets the curve, the slope function of the curve must be equal to the slope of the
tangent, which we know to be negative one. So equating these two expressions
gives the equation negative π₯ over π¦ equals negative one. Multiplying both sides of this
equation by negative π¦ gives π₯ equals π¦.

We now know that at the point where
the tangent meets the curve, the π₯- and π¦-coordinates are the same. We can therefore replace π¦ squared
with π₯ squared in the equation of the curve to give an equation in π₯ only, which
will enable us to find the π₯-coordinate of this point. Solving this equation, we find that
π₯ squared is equal to 36. π₯ is therefore equal to the square
root of 36, taking only the positive solution here as the point lies in the first
quadrant, and so we know π₯ is positive. We find then that the π₯-coordinate
of this point is six.

As weβve determined that the π₯-
and π¦-coordinates are equal at this point, it follows that the π¦-coordinate is
also equal to six.

Weβve now determined the slope of
the tangent and the coordinates of one point that lies on it, and so weβre ready to
find its equation. Using the pointβslope form of the
equation of a straight line and substituting negative one for the slope π and the
point six, six for π₯ one, π¦ one, we find that the equation of the line is π¦ minus
six equals negative one multiplied by π₯ minus six. Distributing the negative one over
the parentheses on the right-hand side and then collecting all terms on the
left-hand side of the equation, we find that the equation of the tangent is π₯ plus
π¦ minus 12 equals zero.