\(\frac{9}{2}\)
\(\frac{3\sqrt17}{2}\)
\(\frac{3\sqrt17}{4}\)
9
Let point P be (h, k)
\((ℎ–1)^2+(k–2)^2+(ℎ+2)^2+(k–1)^2=14\)
\(2ℎ^2+2k^2+2ℎ–6k–4=0\)
Locus of point P : x2 + y2 + x – 3y – 2 = 0
Intersection with x-axis,
x2 + x – 2 = 0
x = –2, 1
Intersection with y-axis,
y2 – 3y – 2 = 0
\(y=\frac{3±\sqrt17}{2}\)
Area of the quadrilateral ACBD is \(=\frac{1}{2}(|x1|+|x2|)(|y1|+|y2|)\)
\(=\frac{1}{2}×3×\sqrt17=\frac{3\sqrt17}{2}\)
So, the correct option is (B): \(\frac{3\sqrt17}{2}\)
Let $ S $ denote the locus of the point of intersection of the pair of lines $$ 4x - 3y = 12\alpha,\quad 4\alpha x + 3\alpha y = 12, $$ where $ \alpha $ varies over the set of non-zero real numbers. Let $ T $ be the tangent to $ S $ passing through the points $ (p, 0) $ and $ (0, q) $, $ q > 0 $, and parallel to the line $ 4x - \frac{3}{\sqrt{2}} y = 0 $.
Then the value of $ pq $ is
Let $ \mathbb{R} $ denote the set of all real numbers. Then the area of the region $$ \left\{ (x, y) \in \mathbb{R} \times \mathbb{R} : x > 0, y > \frac{1}{x},\ 5x - 4y - 1 > 0,\ 4x + 4y - 17 < 0 \right\} $$ is
If \[ f(x) = \int \frac{1}{x^{1/4} (1 + x^{1/4})} \, dx, \quad f(0) = -6 \], then f(1) is equal to: