(a) The maximum and minimum distances of the Earth from the Sun respectively are $152\times10^6$ km and $94.5\times10^6$ km. The Sun is at one focus of the elliptical orbit. Show that the distance from the Sun to the other focus is $575\times10^5$ km. **OR** (b) Prove by vector method $\sin(A+B)=\sin A\cos B+\cos A\sin B$.

> [!TLDR] > (a) uses the ellipse property that aphelion and perihelion distances give $a+c$ and $a-c$ to find the inter-focal distance $2c$; (b) proves the sine addition formula by the vector (cross-product) method. > > **(a) Distance between the two foci of Earth's orbit** > 1. Let the Sun sit at one focus $S$ of the elliptical orbit, with semi-major axis $a$ and focal distance $c$ (centre to focus). > 2. Maximum distance from the Sun (aphelion) $=a+c=152\times10^6$ km. > 3. Minimum distance from the Sun (perihelion) $=a-c=94.5\times10^6$ km. > 4. Subtract: $(a+c)-(a-c)=2c=152\times10^6-94.5\times10^6=57.5\times10^6$ km. > 5. The distance between the two foci is $2c=57.5\times10^6$ km $=575\times10^5$ km — this is the distance from the Sun (one focus) to the other focus. > > **(b) Vector proof of sin(A+B)** > 1. Let $\hat a=\cos A\,\hat i+\sin A\,\hat j$ (unit vector at angle $A$ to the x-axis) and $\hat b=\cos B\,\hat i-\sin B\,\hat j$ (unit vector at angle $-B$ to the x-axis). …

(a) The maximum and minimum distances of the Earth from the Sun respectively are 152×10⁶ km and 94.5×10⁶ km. T

Question 110 of 126

152\times10^6

(b) Prove by vector method

\sin(A+B)=\sin A\cos B+\cos A\sin B

Puducherry TnboardTamil Nadu HSC (DGE) Board 2022Subjective· 5mImportance★★★★★

87% · 110/126 Questions

🔒 Locked · start free trial →

You're viewing a preview — the full solution, concept, methods & PYQ mapping are locked.

Start your 14-day free trial to unlock the full solution →

(a) uses the ellipse property that aphelion and perihelion distances give $a+c$ and $a-c$ to find the inter-focal distance $2c$ ; (b) proves the sine addition formula by the vector (cross-product) method.

(a) Distance between the two foci of Earth's orbit

Let the Sun sit at one focus $S$ of the elliptical orbit, with semi-major axis $a$ and focal distance $c$ (centre to focus).
Maximum distance from the Sun (aphelion) $=a+c=152\times10^6$ km.
Minimum distance from the Sun (perihelion) $=a-c=94.5\times10^6$ km.
Subtract: $(a+c)-(a-c)=2c=152\times10^6-94.5\times10^6=57.5\times10^6$ km.
The distance between the two foci is $2c=57.5\times10^6$ km $=575\times10^5$ km — this is the distance from the Sun (one focus) to the other focus.

(b) Vector proof of sin(A+B)

Let $\hat a=\cos A\,\hat i+\sin A\,\hat j$ (unit vector at angle $A$ to the x-axis) and $\hat b=\cos B\,\hat i-\sin B\,\hat j$ (unit vector at angle $-B$ to the x-axis). …

Start your 14-day free trial to unlock the full solution →

←Previous questionQ40 — (a) Show that the equation of the parabola with focus (-√2, 0) and dir…→Next questionQ42 — The number of normals that can be drawn from a point to the parabola y…

Q.(a) The maximum and minimum distances of the Earth from the Sun respectively are $152\times10^6$ km and $94.5\times10^6$ km. The Sun is at one focus of the elliptical orbit. Show that the distance from the Sun to the other focus is $575\times10^5$ km. OR

Concept understanding — Real-life Applications of Conics

🎓Unlock everything free for 14 days

Unlock everything free for 14 days