Super-Hard SAT Advanced Math Question 18 | Free SAT Prep

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Question 18·25 Super-Hard SAT Advanced Math Questions·Advanced Math

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In the $xy$ -plane, the distinct points $P$ and $Q$ are the solutions to the system below, where $k$ is a constant.

\begin{aligned} x^2+y^2-8x+2ky+7&=0 \\ y&=x+1 \end{aligned}

Both $P$ and $Q$ have positive $y$ -coordinates, and the distance between $P$ and $Q$ is $2\sqrt{2}$ . What is the value of $k$ ?

Your answer

When a system consists of a line and a quadratic curve, substitute the line equation to create a one-variable quadratic. If the question describes two intersection points, interpret its geometric information as information about the roots. Here, the slope of the line converts the distance between points into the difference between their $x$ -coordinates, and the root difference can be found efficiently from the discriminant.

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Compare horizontal and vertical changes

Because both points are on $y=x+1$ , their changes in $x$ and $y$ have equal magnitude. Use this fact with the distance formula.

Turn the system into one quadratic

Replace $y$ with $x+1$ in the first equation. Its two roots represent the $x$ -coordinates of the intersection points.

Connect root separation to the discriminant

For a quadratic with leading coefficient $1$ , the square of the difference of its roots equals its discriminant. Then use the condition about positive $y$ -coordinates to choose between the possible values of $k$ .

Desmos Guide

Find the possible parameter values

Algebra is faster for this question, but Desmos can verify the parameter values. Graph $y=(x-3)^2-4(x+4)$ and $y=4$ . The $x$ -coordinates of their intersections are the possible values of $k$ .

Check the coordinate restriction

Enter $x^2+y^2-8x+2ky+7=0$ and $y=x+1$ , and create a slider for $k$ . Set the slider to each candidate value. Check the intersection points: only one setting gives two points with positive $y$ -coordinates.

Step-by-step Explanation

Use the line to interpret the distance

Since both points lie on $y=x+1$ , a change of $\Delta x$ between the points produces the same change of $\Delta y$ .

Thus,

(2\sqrt{2})^2=(\Delta x)^2+(\Delta y)^2=2(\Delta x)^2

so $|\Delta x|=2$ .

Substitute the line equation

Substitute $y=x+1$ into the circle equation:

\begin{aligned} x^2+(x+1)^2-8x+2k(x+1)+7&=0 \\ x^2+(k-3)x+(k+4)&=0 \end{aligned}

The two roots of this quadratic are the $x$ -coordinates of $P$ and $Q$ .

Use the separation of the roots

For a monic quadratic $x^2+bx+c=0$ , the square of the difference between its roots is $b^2-4c$ .

Here, the roots differ by $2$ , so

\begin{aligned} (k-3)^2-4(k+4)&=2^2 \\ k^2-10k-11&=0 \\ (k-11)(k+1)&=0 \end{aligned}

Therefore, $k$ could be $11$ or $-1$ .

Apply the positive-coordinate condition

If $k=11$ , the quadratic is

x^2+8x+15=0

with roots $-3$ and $-5$ . The corresponding $y$ -coordinates are $-2$ and $-4$ , which are not positive.

If $k=-1$ , the roots are $1$ and $3$ , giving positive $y$ -coordinates $2$ and $4$ . Therefore, $k=-1$ .

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Desmos Guide

Step-by-step Explanation

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Strategy

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Step-by-step Explanation