JEE MAIN - Mathematics (2019 - 9th January Morning Slot - No. 22)
If $$\theta $$ denotes the acute angle between the curves, y = 10 – x2 and y = 2 + x2 at a point of their intersection, the |tan $$\theta $$| is equal to :
$$8 \over 15$$
$$4 \over 9$$
$$7 \over 17$$
$$8 \over 17$$
Explanation
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Angle between the curves is the acute angle between the tangents at the point of intersection.
y = 10 $$-$$ x2 (for curve 1)
and y = 2 + x2 (for curve 2)
$$ \therefore $$ 10 $$-$$ x2 = 2 + x2
$$ \Rightarrow $$ 2x2 = 8
$$ \Rightarrow $$ x2 = 4
$$ \Rightarrow $$ x = 2, $$-$$ 2
$$ \therefore $$ points of intersection (2, 6) and ($$-$$ 2, 6)
$${{dy} \over {dx}}$$ for curve 1 = $$-$$ 2x
$$ \therefore $$ Slope(m1) of curve 1 is = $$-$$ 2(2) = $$-$$ 4
$${{dy} \over {dx}}$$ for curve 2 = 2x
$$ \therefore $$ slope (m2) of curve 2 = 2 $$ \times $$ 2 = 4
$$ \therefore $$ tan$$\theta $$ = $$\left| {{{{m_1} - {m_2}} \over {1 + {m_1}{m_2}}}} \right|$$
= $$\left| {{{ - 4 - 4} \over {1 + \left( { - 16} \right)}}} \right|$$
= $${8 \over {15}}$$
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