Question

find the equation to the pair of lines passing through the origin and perpendicular to 5x²-2xy-3y² = 0 ​

Answer 1

Answer:

Step-by-step explanation:

0

Answer 2

Answer:

Step-by-step explanation:

So, you have a general 2nd degree equation as-

ax2+by2+2hxy+2gx+2fy+c=0

As far as we're concerned, it could represent a circle, an ellipse, a parabola, a hyperbola or a pair of straight lines.

We need to find a relation between x and y to find out what it represents. So, treat the equation as a quadratic in one variable.

ax2+(2hy+2g)x+(by2+2fy+c)=0

Correlate this with a general quadratic equation,

AX2+BX+C=0

And use the Quadratic Formula-

x=−2hy−2g2a±(2hy+2g)2–4a(by2+2fy+c)−−−−−−−−−−−−−−−−−−−−−−−−√2a

⇒x=−hy−ga±h2y2+g2+2hgy−aby2–2afy−ac−−−−−−−−−−−−−−−−−−−−−−−−−−−−√a

Now, the term inside the square root needs to be a perfect square, only then can we have a linear relation between x and y. As you can see, the term there also happens to be a quadratic, but in y, and so we set it's determinant to zero.

Examining this term a bit closely, we get-

(h2−ab)y2+2(hg−af)y+(g2−ac)=0

Set the determinant to zero as discussed above:

(hg−af)2=(h2−ab)(g2−ac)

Expand it, and you'll see that:

abc+2gfh−af2−bg2−ch2=0,a≠0

So, we conclude, as long as this condition is satisfied, the equation represents a pair of straight lines.

Now that the term inside the square root is a perfect square, it's root can be found as follows-

(h2−ab)y2+2(hg−af)y+(g2−ac)=0

⇒y=−2(hg−af)2(h2−ab)

So the equation reduces to-

(y+hg−afh2−ab)2=0

Plug that into the equation in x-

x=−hy−g±(y+hg−afh2−ab)a

Well, these are the equations you needed. You just need to expand them a bit further, which I'm not doing here.

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