Tangents and Normals of rectangular hyperbolas

Question

Please, could someone explain the solution to (d)

I solved (a), (b) and (c) however, I don't understand how to calculate (d).

(a) displayed in the question
(b) $ q^2 y + x = 10q $
(c) displayed in the question
(d) ?

I'm not sure what $p^2 q^2$ even infers.

P.s. this is just personal/curiosity study, not, homework or classwork etc.

Answer 1

Truth of matter is that we're not actually told what to do, nor is there a method to be sure of what to do -- at all. I'm sure that even the publisher would state that "doing this and that" will hopefully give us the correct solution to whatever $p^2 q^2$ is. A.k.a, telepathy dependent questions. The equation determines the foci i presume - maybe even the eccentricity.

Here goes;

1. Calculate the gradient of PQ...

$$=\left(\frac{\frac{5}{p}-\frac{5}{q}}{5p-5q}\right)$$ $$=\left(\frac{\frac{5q}{pq}-\frac{5p}{pq}}{5p-5q}\right)$$ $$\left(\frac{\frac{b}{c}}{a}\right) = \left(\frac{b}{(c).(a)}\right)=\left(\frac{5q-5p}{pq(5p-5q)}\right)$$ $$=\left(\frac{-(5p-5q)}{pq(5p-5q)}\right)$$

• $(5p-5q)$ is a common factor so it cancels to;
  $$-\frac{1}{pq}$$

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2. Calculate the gradient of ON...

$$=\left(\frac{\frac{10}{p+q}-0} {\frac{10pq}{p+q}-0}\right)$$

• "10" and "p+q" both cancel out due to the expressions being common factor across the top and bottom side of the fraction.

$$\frac{1}{pq}$$

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3. Multiply the two gradients to solve...

$$\left(-\frac{1}{pq}\right).\left(\frac{1}{pq}\right)$$

$$=-\frac{1}{p^2 q^2}$$

• When the two lines intersect in a specific manner such that they create 4, 90° angles and hence, are known as perpendicular. As a result to this, we know that the product of the multiplication is "equal to", -1. i.e. $m_1 m_2 = -1$. (only applicable when the gradients in context are of 2 perpendicular lines)

Therefore;

$$-\frac{1}{p^2 q^2} = -1$$

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4. Rearrange and simplify...

$$-1 = -1(p^2 q^2)$$ $$1 = p^2 q^2$$

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5. Final answer...

$$p^2 q^2 = 1$$

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6. End note...

$p=\sqrt{\frac{1}{q^2}}$ or $p=-\sqrt{\frac{1}{q^2}}$

$q=\sqrt{\frac{1}{p^2}}$ or $q=-\sqrt{\frac{1}{p^2}}$

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p.s. yes, i solved my own question.

Answer 2

You can find the slope of line $ON$ from point (c) and the slope of line $PQ$: since those two lines are perpendicular, the product of their slopes must be $-1$. From that you get $p^2q^2=1$.