Two Dimensional Analytical Geometry-II — Book-Back Answer Key (Exercise 5.6)
Two Dimensional Analytical Geometry-II — Book-Back Answer Key (Exercise 5.6)
Verified answers to all 25 multiple-choice questions of Exercise 5.6, each with a one-line justification. Options are numbered as printed: (1)–(4).
Correction to the draft key: Question 13 (the inscribed–circumscribed ellipse problem) works out to eccentricity \(e=\tfrac{1}{2}\), which is option (3). The supplied draft key listed option 4 (\(\tfrac{3}{4}\)); that is a slip and the value \(\tfrac{1}{2}\) is used here.
| Q | Correct option | Reason |
|---|---|---|
| 1 | (1) \(0,\ -\dfrac{40}{9}\) | Both the base circle and the chord vanish at the two given points; tangency to the \(y\)-axis (\(f^{2}=c\)) forces \(\lambda=0\) or \(-\tfrac{40}{9}\). |
| 2 | (3) \(\dfrac{2}{\sqrt{3}}\) | From \(\tfrac{2b^{2}}{a}=8\) and \(2b=c\) with \(c^{2}=a^{2}+b^{2}\), the eccentricity is \(e=\tfrac{c}{a}=\tfrac{2}{\sqrt{3}}\). |
| 3 | (4) \(-35| Centre \((2,4)\), radius \(5\); two intersections need the centre-to-line distance below \(5\), i.e. \(-35 | |
| 4 | (3) \(\dfrac{10}{3}\) | Centre \((1,k)\) with radius \(|k|\); passing through \((2,3)\) gives \(k=\tfrac{5}{3}\), so the diameter is \(\tfrac{10}{3}\). |
| 5 | (3) \(\sqrt{10}\) | Equal \(x^{2}\) and \(y^{2}\) coefficients force \(b=3\); the reduced circle has radius \(\sqrt{10}\). |
| 6 | (1) \((4,7)\) | The sides are \(x=2,6\) and \(y=5,9\), a square whose inscribed circle is centred at \((4,7)\). |
| 7 | (1) \(x+2y=3\) | A normal passes through the centre \((1,1)\); with slope \(-\tfrac{1}{2}\) it is \(x+2y=3\). |
| 8 | (3) \(8\) | The sum of focal distances equals the major axis \(2a=8\) since \(a=4\). |
| 9 | (2) \(2\sqrt{5}\) | The centre is the intersection \((8,-2)\) of the two diameters; its distance to \((6,2)\) is \(2\sqrt{5}\). |
| 10 | (2) \(2(a^{2}+b^{2})\) | The four foci form a rhombus with equal diagonals \(2\sqrt{a^{2}+b^{2}}\), so the area is \(2(a^{2}+b^{2})\). |
| 11 | (1) \(\sqrt{2}\) | The latus-rectum normals \(x\pm y=3\) are tangent to the circle at \((3,-2)\); the distance is \(\sqrt{2}\). |
| 12 | (4) \(9\) | With \(a=3\) and slope \(-1\), the normal is \(x+y=9\), so \(k=9\). |
| 13 | (3) \(\dfrac{1}{2}\) | \(E_{2}:\,\tfrac{x^{2}}{12}+\tfrac{y^{2}}{16}=1\) (vertical major axis) gives \(e=\tfrac{1}{2}\) — this corrects the draft key, which listed option 4 (\(\tfrac{3}{4}\)). |
| 14 | (3) \(\left(\dfrac{9}{2\sqrt{2}},\,\dfrac{1}{\sqrt{2}}\right)\) | Slope-\(2\) tangency gives \(y_{1}^{2}=\tfrac{1}{2}\); one contact point is \(\left(\tfrac{9}{2\sqrt{2}},\tfrac{1}{\sqrt{2}}\right)\). |
| 15 | (1) \(x^{2}+y^{2}-6y-7=0\) | Foci \((\pm\sqrt{7},0)\); a circle centred \((0,3)\) with \(r^{2}=16\) is \(x^{2}+y^{2}-6y-7=0\). |
| 16 | (4) \(\dfrac{1}{4}\) | \(T=(0,r)\) through the origin; external tangency to \(C\) gives \(r=\tfrac{1}{4}\). |
| 17 | (4) \(40\) | Here \(a=5,\ b=4\); the rhombus area is \(\tfrac{1}{2}(10)(8)=40\). |
| 18 | (1) \(2ab\) | The largest inscribed rectangle has area \(2ab\), reached at \(\theta=\tfrac{\pi}{4}\). |
| 19 | (1) \(\dfrac{1}{\sqrt{2}}\) | A right angle at \(B\) gives \(c^{2}=b^{2}\), hence \(e=\tfrac{1}{\sqrt{2}}\). |
| 20 | (2) \(\dfrac{1}{3}\) | The relation is \(PF^{2}=\tfrac{1}{9}PM^{2}\), so \(e=\tfrac{1}{3}\). |
| 21 | (2) \(x=-1\) | The locus of vertices of perpendicular tangents is the directrix \(x=-1\). |
| 22 | (3) \((5,-2)\) | Centre \((3,-2)\), radius \(2\); the circle also passes through \((5,-2)\). |
| 23 | (3) \(\dfrac{x^{2}}{9}+\dfrac{y^{2}}{5}=1\) | The focus–directrix law with \(e=\tfrac{2}{3}\) yields the ellipse \(\tfrac{x^{2}}{9}+\tfrac{y^{2}}{5}=1\). |
| 24 | (3) \(0\) | Tangency gives \(m=\pm2\); the roots of the quadratic sum to \(a+b=0\). |
| 25 | (2) \((-3,2)\) | The centre \((4,2)\) is the midpoint of the diameter, so the other end is \((-3,2)\). |